Copyright © 2023 Shabbir Basrai  
Revision Date: March 2023  
Contents  
1
Why Treat Wastewater? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13  
1.1  
1.2  
1.3  
13  
13  
13  
2
2.1  
Regulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15  
2.1.1 Treated Wastewater - NPDES Permit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15  
2.1.2 Influent Wastewater - National Pretreatment Program . . . . . . . . . . . . . . . . . . . 15  
2.2  
2.3  
2.4  
16  
17  
17  
2.4.1 Operator Certification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17  
2.4.2 Worker Safety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22  
3
Elements of Wastewater Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23  
3.1  
3.2  
3.3  
23  
23  
24  
3.3.1 Liquid Phase Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24  
3.3.2 Treatment of Wastewater Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24  
3.4  
26  
4
Constituents Properties and Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 27  
4.1  
4.1.1 Organics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27  
4.1.2 Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29  
4.1.3 Nutrients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31  
4.2  
33  
4.2.1 pH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33  
4.2.2 Oxidation Reduction Potential (ORP) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34  
4.2.3 Chlorine Residual . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35  
4.2.4 Alkalinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35  
4.2.5 Dissolved Oxygen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36  
4.2.6 Microbiological testing and monitoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36  
4.2.7 Specific Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38  
4.3  
39  
4.3.1 Sampling Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39  
4.3.2 Data Reporting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40  
4.4  
41  
4.4.1 BOD Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41  
4.4.2 Wastewater solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42  
4.4.3 Bacteriological Enumeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46  
5
Collections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75  
5.1  
75  
75  
5.2  
5.2.1 Storm-water systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75  
5.2.2 Combined sewer systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75  
5.3  
5.4  
76  
76  
5.4.1 Infiltration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77  
5.4.2 Inflow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77  
5.4.3 Odors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77  
5.4.4 Fats, Oils and Grease (FOG) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77  
6
Preliminary Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83  
6.1  
6.1.1 Screening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83  
6.1.2 Grinding and Shredding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84  
6.1.3 Flow Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84  
6.1.4 Grit Removal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87  
6.1.5 Flow Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88  
6.1.6 Pre-aeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88  
6.1.7 Flow Equalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89  
7
Primary Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103  
7.1  
7.1.1 Inlet Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105  
7.1.2 Settling Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105  
7.1.3 Sludge Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105  
7.1.4 Skimming Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106  
7.1.5 Outlet Zone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107  
7.2  
7.3  
7.4  
109  
109  
109  
7.4.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109  
7.4.2 Process/mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110  
8
Introduction to Secondary Treatment . . . . . . . . . . . . . . . . . . . . . . . . . 123  
8.1  
8.2  
8.3  
123  
124  
124  
9
Trickling Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125  
9.1  
9.2  
9.3  
9.4  
9.5  
125  
126  
127  
127  
127  
9.5.1 Low-rate filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127  
9.5.2 High-rate filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127  
9.5.3 Roughing filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128  
9.6  
128  
9.6.1 Ponding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128  
9.6.2 Odors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128  
9.6.3 Trickling Filter Flies - Psychoda . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128  
9.6.4 Cold weather problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128  
10  
Stabilzation Ponds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139  
139  
139  
139  
10.3.1 Anaerobic Ponds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139  
10.3.2 Facultative Ponds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140  
10.3.3 Aerobic Stabilization Ponds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140  
142  
11  
Activated Sludge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163  
163  
164  
11.2.1 Dissolved oxygen (DO) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164  
11.2.2 pH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164  
11.2.3 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164  
11.2.5 Food to Mass ratio (F:M) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165  
11.2.6 Sludge age . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165  
11.2.7 Sludge settleability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165  
166  
168  
11.4.1 Controlling waste activated sludge (WAS) pumping rate . . . . . . . . . . . . . . . . 168  
11.4.2 Controlling the return activated sludge (RAS) rate . . . . . . . . . . . . . . . . . . . . . 169  
170  
173  
11.6.1 Conventional activated sludge process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173  
11.6.2 Conventional step feed process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173  
11.6.3 Contact stabilization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174  
11.6.4 Pure oxygen system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175  
11.6.5 Sequencing batch reactors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176  
11.6.6 Extended aeration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177  
11.6.7 Krauss process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178  
12  
Nutrient Removal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191  
191  
191  
12.2.1 Nitrogen Removal Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192  
200  
12.3.1 Phosphorous removal methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200  
13  
Solids Treatment Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215  
215  
215  
216  
216  
14  
Biosolids Regulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217  
217  
217  
217  
218  
14.4.1 Pollutant concentration limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218  
14.4.2 Pathogen reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218  
14.4.3 Vector attraction reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219  
15  
Sludge Thickening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225  
225  
225  
225  
15.3.1 Principles of gravity thickener operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226  
15.3.2 Elements of a gravity thickener . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227  
15.3.3 Gravity thickener operational parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . 227  
227  
15.4.1 Principles of DAFT operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228  
15.4.2 Elements of a DAFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229  
15.4.3 DAFT operational parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229  
16  
Solids Stabilization - Digestion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235  
236  
236  
239  
16.3.1 Digestion temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239  
16.3.2 Digestion mixing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239  
16.3.3 Digestion feed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239  
16.3.4 Volatile solids breakdown . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240  
16.3.5 Digester pH and alkalinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241  
16.3.6 Digester pH control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242  
16.3.7 Digester gas production . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242  
16.3.8 Digester failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243  
16.3.9 Digester toxicity issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243  
16.3.10 Nutrients and trace constituents requirement . . . . . . . . . . . . . . . . . . . . . . . . . 243  
243  
17  
Solids Dewatering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257  
17.1.1 Principles of belt filter press operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258  
17.1.2 Elements of the belt filter press . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259  
17.1.3 Belt press operational parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260  
260  
17.2.1 Principles of centrifuge operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260  
17.2.2 Elements of the Centrifuge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261  
17.2.3 Centrifuge operational parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262  
18  
Disinfection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265  
265  
266  
18.2.1 Chlorine properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266  
18.2.2 Chlorine storage and safety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267  
18.2.3 Forms of chlorine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268  
18.2.4 Chlorine reactions related to disinfection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268  
18.2.5 Chlorine disinfection process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269  
18.2.6 Factors affecting chlorine disinfection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269  
18.2.7 Chlorine Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269  
18.2.8 Chlorination Byproducts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 270  
18.2.9 Chloroamination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271  
18.2.10 Chlorine dosing terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273  
18.2.11 Chlorine dosing problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273  
19  
Odor Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297  
297  
297  
298  
298  
19.4.1 Hydrogen Sulfide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300  
19.4.2 Ammonia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300  
300  
300  
304  
19.7.1 Common Vapor Phase Odor Control Methods . . . . . . . . . . . . . . . . . . . . . . . . 304  
20  
Wastewater Chemicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315  
316  
317  
20.2.1 Polymers in wastewater treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317  
20.3.1 lbs chemicals needed given flow and dosing rate . . . . . . . . . . . . . . . . . . . . . 318  
20.3.2 Chemical batching and dilution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318  
21  
Pumping Efficiencies and Costs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325  
325  
328  
329  
330  
333  
22  
Water Math . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347  
347  
348  
348  
350  
350  
352  
352  
22.7.1 Percentage Concentrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353  
354  
356  
357  
357  
357  
358  
358  
360  
22.15.1 Example Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361  
361  
363  
364  
22.18.1 Chemical Dosing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364  
22.18.2 Chemical Dilution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365  
366  
366  
22.20.1 Channel Velocity and Flow Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366  
22.20.2 Grit Removal Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367  
367  
22.21.1 Hydraulic or Surface Loading Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367  
22.21.2 Detention Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367  
22.21.3 Weir Overflow Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368  
22.21.4 Removal Efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368  
22.21.5 Solids Removal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369  
370  
22.22.1 Hydraulic or surface loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370  
22.22.2 BOD and TSS Removal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370  
22.22.3 Organic Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370  
22.22.4 Recirculation Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371  
371  
22.23.1 Pond Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371  
22.23.2 Solids Loading Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371  
22.23.3 Organic Loading Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372  
22.23.4 Detention time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372  
22.23.5 Hydraulic Loading Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372  
373  
22.24.1 Mean Cell Residence Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373  
22.24.2 F:M (Food to Microorganism Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374  
22.24.3 Sludge Volume Index (SVI) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375  
22.24.4 Sample Math Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376  
377  
378  
22.26.1 Sludge Pumping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378  
22.26.2 Sludge Blending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378  
22.26.3 Digester VS or Organic Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379  
22.26.4 Total volatile solids (VS) reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380  
22.27.1 Calculation of solids recovery . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381  
22.27.2 Calculation of dewatered cake volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381  
22.27.3 Hauling cost impact due to solids content change . . . . . . . . . . . . . . . . . . . . 381  
23  
Wastewater Treatment - Future . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475  
23.1.1 Water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475  
23.1.2 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476  
23.1.3 Nutrients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476  
476  
23.2.1 Constituents of Emerging Concern . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476  
23.2.2 Decentralized and Distributed Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477  
23.2.3 Climate Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477  
24  
25  
Wastewater Treatment - Careers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479  
Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481  
1. Why Treat Wastewater?  
1.1 Definition of Wastewater  
Wastewater is human-polluted water from home and industries. This includes water from:  
• Flushing toilets and urinals - blackwater.  
• Bathing, showering, and washing clothes and dishes - greywater.  
• Commercial and industrial activities.  
...and often included is stormwater which contain pollutants washed off from inhabited  
areas - roads, parking lots, and rooftops.  
1.2 Why Treat Wastewater  
Although nature has an inherent capability to breakdown pollutants, given the large quantity gen-  
erated from human activities, there is a need for centralized wastewater treatment plants to treat  
wastewater before releasing it back to the environment. Wastewater from homes, businesses, and  
industries are collected in sewers for delivery to the treatment plant and subsequently discharged  
to a water body like a lake, river or ocean, or land, or reused.  
Wastewater treatment is designed to remove:  
• organic matter  
• inorganic pollutants including plant nutrients - nitrogen and phosphorous  
• pathogenic (disease causing) organisms  
1.3 Benefits of Treating Wastewater  
Wastewater treatment protects:  
• The environment  
 
 
 
 
14  
Chapter 1. Why Treat Wastewater?  
• Human health  
Specifically, wastewater treatment allows for the following:  
1. Mitigates deterioration of the receiving waters’ ecosystem  
The discharge of inadequately treated wastewater will cause oxygen levels in the receiving  
waters to be depleted, due to:  
Wastewater containing nitrogen and phosphorus based pollutants (plant nutrients)  
entering a water body such as a lake or river will promote plant and algae growth  
which will seriously impact its normal aquatic life including fish through a process  
similar to the following:  
Nutrient promote algae bloom  
Algae bloom prevent sunlight to the native plant spieces below the water’s sur-  
face causing native plants to die  
The organic material from the dead plants and algae promote growth of aerobic  
bacteria which will consume the dissolved oxygen in the water resulting in  
oxygen depletion.  
The natural aquatic life including fish, frogs, and turtles will not be able to sur-  
vive under oxygen depleted conditions and will die or leave that zone.  
Other organic material present in wastewater will similarly promote growth of aero-  
bic bacteria, intensifying the eutrophication of the receiving waters.  
Thus, proper treatment of wastewater will prevent eutrophication - which is the depletion  
of dissolved oxygen of the receiving water, consequently impacting/destroying its normal  
aquatic life.  
2. Removal of other harmful pollutants  
Organic and inorganic pollutants, including metals such as mercury, lead, cadmium,  
chromium and arsenic can have acute and chronic toxic effects on aquatic species and  
wildlife including migratory birds, are removed during the wastewater treatment process.  
3. Removal of pathogens  
Wastewater treatment removes parasites and disease-causing pathogens including bacteria  
and viruses for:  
People to continue enjoying recreational activities in the receiving bodies of waters  
such as lakes and rivers  
Preventing the contamination of fish and other consumable products obtained from  
the waters  
• Allow the water body to remain as the source of potable water  
4. Reclaim water for recycle or reuse  
Besides protecting human health and the environment, wastewater treatment paves way  
for establishing the reuse or recycle of treated wastewater. This benefit is particularly  
important for densely populated areas with limited access to fresh water supplies.  
2. Regulations  
2.1 Regulations Related to Wastewater Treatment  
The main objective of these regulations is to ensure appropriate quality of the treated wastewater.  
2.1.1 Treated Wastewater - NPDES Permit  
The National Pollutant Discharge Elimination System (NPDES) permit program program ad-  
dresses water pollution by regulating point sources that discharge pollutants to waters of the  
United States.  
The NPDES permit program was created in 1972 by the Clean Water Act (CWA)and is  
administered by the federal USEPA.  
• Applies to sources that discharge pollutants to waters of the United States.  
Requires all facilities discharging “pollutants” into any body of water in the USA to obtain  
and comply with a NPDES permit.  
• NPDES permit establishes discharge limits, monitoring and reporting requirements  
In California, the responsibility of implementing the federal NPDES program is delegated  
to the State of California through the State Water Resources Control Board (State Water  
Board or SWRCB) and finally to the nine Regional Water Quality Control Boards (Re-  
gional Water Boards or RWQCB), collectively known as Water Boards.  
• The RWQCB issues the NPDES permit.  
2.1.2 Influent Wastewater - National Pretreatment Program  
Municipal wastewater treatment plants also known as Publicly Owned Treatment Works (POTWs)  
implement and enforce their Pretreatment or Industrial Discharge Control programs to meet Fed-  
eral and State regulations requirements related to wastewater discharges from industrial sources.  
 
 
 
 
16  
Chapter 2. Regulations  
The national pretreatment program is a component of the NPDES program and is also known as  
the Source Control Program  
The Pretreatment/Source Control program is for controlling industrial (non-domestic) wastewater  
discharges with the following objectives:  
1. Protect the treatment plant operations so that the industrial discharge does not contain pol-  
lutants or have certain characteristics (including pH, temperature) which could adversely  
effect the treatment process or impact public safety and the safety of the people working at  
the treatment plant.  
2. Prevent the introduction of pollutants that could pass through untreated and into the receiv-  
ing body of water.  
3. Improve opportunities for reuse or recycling of wastewater and sewage sludge.  
In California:  
1. Wastewater treatment plants are required to have a Pretreatment Program when their total  
design flows are greater than five million gallons per day (5 mgd).  
2. Facilities with smaller flows (5 mgd or less) may also be required to implement a Pretreat-  
ment Program if they receive industrial waste and pretreatment is warranted.  
3. The Pretreatment Program for a wastewater treatment entity is reviewed and approved by  
the State and Regional Water Boards, and  
4. The Pretreatment Program’s monitoring and reporting requirements are incorporated in the  
facility’s NPDES permit.  
2.2 Sewage Sludge/Biosolids Regulations  
Sewage sludge or biosolids is a byproduct of wastewater treatment. The biosolids produced  
are disposed or used using methods including land application, landfill and incineration. Fed-  
eral Regulation 40CFR Part 503 also known as Rule 503 establishes the standards for the use  
or disposal of wastewater biosolids - as stipulated under the Clean Water Act. The facility’s  
NPDES permit incorporates the applicable federal, state and local requirements as they apply to  
its biosolids.  
Part 503 rule applies to any person who applies biosolids to the land or fires biosolids in a  
biosolids incinerator, and to the owner/operator of a surface disposal site, or to any person  
who is a preparer or generator of biosolids for use, incineration, or disposal.  
• Part 503 standard includes:  
1. General requirements which establishes the purpose and applicability of the rule, the  
compliance period, and exclusions from the rule.  
2. Limits on heavy metals content  
3. Solids management practices related to use and disposal of wastewater biosolids  
4. Operational standards related to biosolids management, and  
5. Requirements for the frequency of monitoring, record-keeping, and reporting  
Part 503 requirements are factored in when establishing the heavy metals concentration limits for  
 
2.3 Air Quality Regulations  
17  
the Pretreatment or Industrial Control Program as a significant portion of the heavy metals in the  
influent wastewater are removed as part of the wastewater solids.  
2.3 Air Quality Regulations  
Air emissions from wastewater collections and treatment systems are subject to federal,  
state and local air quality related rules and regulations established to protect human health  
and comfort, and the environment.  
Typically, a local agency such as the South Coast Air Quality Management District (SCAQMD)  
is designated to enact and enforce air quality rules and regulations, through its permitting  
process, applicable to all sources of air emissions including wastewater treatment plants.  
• Systems/processes subject to air quality regulations at air quality regulations include:  
Fugitive emissions: Foul air containing compounds such as hydrogen sulfide and  
organics, which escape from process tanks, pipes and associated structures such as  
manholes and wetwells, is potentially harmful for the affected public and also cause  
odors.  
Digester gas combustion: Digester gas a product of wastewater solids treatment  
contains methane and is either combusted in power generation equipment or burned  
in flares.  
Odor control systems: These commonly used systems are for controlling emissions  
of regulated pollutants such as ammonia and to prevent odors associated with com-  
pounds such as hydrogen sulfide.  
• Related to its air pollutants emissions, wastewater treatment plants are required to:  
Obtain air quality related operating permits for equipment and processes which emit  
air pollutants and for its systems treating foul air.  
Implement air emission pollutants control measures  
Comply with record keeping and reporting requirements  
Comply with air quality rules to prevent public nuisance and protect public health  
and safety  
2.4 Regulations Related to Operations and Maintenance  
2.4.1 Operator Certification  
The requirements of the Operator Certification program is established for each state. These  
meet the Operator Certification Requirements of the regulations stemming from the 1996  
Amendments to the Safe Drinking Water Act.  
The goal is to ensure that operators of wastewater treatment facilities in the State meet the  
minimum level of competence; thereby, protecting public health and the environment.  
In California, the Wastewater Operator Certification program (WWOCP) administers  
Wastewater Treatment Plant Certification examinations, certifications (grades I to V), and  
certification renewals.  
 
 
 
18  
Chapter 2. Regulations  
WWOCP classifies Wastewater Treatment Plants and stipulates that no person shall oper-  
ate a wastewater treatment plant unless that person has been certified by the division as a  
wastewater treatment plant operator or operator-in-training at a grade appropriate for the  
class of plant being operated.  
A certified operator or operator-in-training may be subject to administrative sanctions  
including reprimand or denial, suspension, probation, or revocation of the operator certifi-  
cation for performing, or allowing or causing another to perform acts which include:  
Operating or allowing the operation of a wastewater treatment plant by a person who  
is not certified at the grade necessary for the position  
failing to use care or good judgment in the course of employment as an operator or  
failing to apply knowledge or ability in the performance of duties.  
Negligence causing the violation of appropriate waste discharge requirements of the  
NPDES permit  
Classification of Wastewater Treatment Plants  
Grade Levels of Operator Certifications  
22  
Chapter 2. Regulations  
2.4.2 Worker Safety  
• Wastewater treatment facility can be an extremely unsafe occupational field  
It involves most of the major categories of workplace hazards: biological, chemical, phys-  
ical, safety and ergonomic, accentuated with other factors such as shift work and diverse  
tasks.  
Entities including The Occupational Safety and Health Administration(OSHA) National  
Electrical Code (NEC), National Fire Protection Association (NFPA), Underwriters Labo-  
ratory (UL) have recognized these hazards and implemented codes and standards to protect  
the affected persons and wastewater workers.  
 
3. Elements of Wastewater Treatment  
Wastewater cycle is a part of the water cycle where the water consumed as part of the normal  
human and industrial activity is returned back to the environment after treatment.  
Wastewater cycle comprises of the following sequential elements:  
1. Generation  
2. Collection  
3. Treatment  
4. Disposal or reuse  
3.1 Generation  
Wastewater originates from domestic, industrial, commercial or agricultural activities. The char-  
acteristics of wastewater vary depending on the source. Types of wastewater include:  
Domestic Sewage: wastewater derived principally from dwellings, business buildings,  
institutions, and  
• Industrial Sewage: liquid waste from industrial processes  
Typical per person generation of wastewater in the USA is about 70-100 gallons per day  
3.2 Collections  
Wastewater is collected from its point of origin - home, businesses, industries etc. and  
conveyed via sewer lines to a centralized wastewater treatment facility.  
When the rainwater drainage is made part of the sewer system, the system is termed as  
Combined System.  
 
 
 
24  
Chapter 3. Elements of Wastewater Treatment  
The system where the sewage is conveyed separately from the stormwater flows is termed  
as Separated System.  
In the Separated System, the Sanitary Sewers convey the wastewater and the Stormwater  
Sewer conveys the storm water flows.  
For the Combined System, rainstorms pose the threat of overwhelming the sewers and the  
treatment plant  
Combined System  
Separated System  
3.3 Treatment  
3.3.1 Liquid Phase Treatment  
Wastewater treatment can involve physical, chemical or biological processes or combina-  
tions of these processes depending on the required outflow standards.  
• Wastewater treatment typically involves a series of steps with increasing level of treatment:  
Preliminary: The preliminary process removes large/coarse solids which include  
rocks, tree branches, grit and other debris present in wastewater.  
Primary: The primary process is also a physical process where the separable wastew-  
ater solids - solids that float and solids that can settle, are removed.  
Secondary: Secondary treatment is a biological treatment process where microorgan-  
isms consume the organic matter present in the wastewater.  
Tertiary or Advanced Treatment: The tertiary/advanced treatment processes improve  
the quality of treated water beyond the secondary treatment level. This process may  
include nutrient removal and disinfection.  
3.3.2 Treatment of Wastewater Solids  
Solids are a byproduct of wastewater treatment.  
Screenings and grit removed as part of the preliminary treatment is typically disposed  
in a landfill.  
Sludge generated from the wastewater treatment processes - settled solids and scum  
from primary and secondary treatment processes needs to be treated prior to disposal  
or reuse to comply with wastewater solids - biosolids regulations.  
 
 
 
3.3 Treatment  
Typical solids treatment is comprised of the following three sequential steps:  
25  
1. Sludge thickening  
2. Sludge stabilization  
3. Sludge dewatering  
Sludge Thickening  
Sludge thickening improves performance of sludge stabilization process and provides  
capital and operational cost savings due to a lower volume of sludge  
Sludge Stabilization  
Sludge stabilization process produces solids (biosolids) that meet Part 503 rule require-  
ments.  
Sludge Dewatering  
Solids stabilized using digestion process has only a small percentage by weight of solids  
-less than 5%. It therefore becomes necessary to dewater the stabilized sludge prior to  
hauling off-site for final disposal. A generalized layout/process sequencing in a wastewater  
treatment plant is shown below:  
Individual wastewater treatment processes involve different process options or sequences  
which are illustrated in the graphic below:  
26  
Chapter 3. Elements of Wastewater Treatment  
3.4 Disposal or Reuse  
Wastewater treatment processes can be designed to dispose the treated water where the  
water is reintroduced to the environment or for reuse where the treated water is reclaimed  
or recycled - for various purposes including irrigation, industrial use or for potable use.  
• Water disposal methods include:  
Surface water discharge  
Subsurface discharge  
• Water reuse methods include:  
Potable water reuse  
Indirect potable reuse: Here the treated water is blended with groundwater or  
*
surface water and then reclaimed and treated further for drinking (potable) water  
use  
Direct potable reuse: Here the treated wastewater is subjected to advanced  
treatment and introduced directly into a municipal water supply system  
*
Water reclamation for irrigation or industrial use  
Land application for beneficial use  
Solids generated from the wastewater treatment process may be removed and disposed to  
a landfill or subject to further treatment which may allow for energy recovery - from the  
organic solids and for beneficial reuse due to its plant nutrient content.  
 
4. Constituents Properties and Analysis  
4.1 Wastewater Constituents  
4.1.1 Organics  
• The main reason for treating domestic wastewater is to remove the organic matter.  
Organics are substances containing carbon, hydrogen and oxygen, and some of which may  
be combined with nitrogen, sulfur or phosphorous.  
About 50 percent of the solids present in wastewater are organic. This fraction is generally  
of animal or vegetable life, dead animal matter, plant tissue or organisms, and also include  
synthetic organic compounds.  
The principal organic compounds present in domestic wastewater are proteins, carbohy-  
drates and fats together with the products of their decomposition.  
Organics are subject to decay or decomposition through the activity of bacteria and other  
living organisms. Since the organic fraction can be driven off at high temperatures, they are  
also called volatile solids.  
Organics in wastewater is typically quantified in terms of oxygen required to oxidize the  
carbon based material present in wastewater using the following methods:  
Biochemical Oxygen Demand (BOD)  
Oxygen is required for the consumption of organic matter by aerobic bacteria  
BOD test measures the depletion of oxygen in a wastewater sample over a five day  
period  
BOD measures the organic content in terms of oxygen required for the microorgan-  
isms to consume the organic material present  
BOD is typically measured as BOD5 which is the oxygen demand of the wastewater  
 
 
 
28  
Chapter 4. Constituents Properties and Analysis  
measured after 5 days of the initiation of the test.  
The test involves incubating a known dilution of wastewater in a 300 ml bottle for 5  
days at 20C. The dissolved oxygen (DO) content at the start and end of the incuba-  
tion period is used for calculating the BOD.  
For the test to be considered valid, the following criteria need to be met: 1) DO con-  
sumption during the test must be at least 2 mg/l, 2) DO remaining at the end of the  
test must be at least 1 mg/l, and 3) DO consumed in blank should be 0.2 mg/l or less  
BOD is a parameter to measure the strength of wastewater and the measurement of  
the wastewater treatment plant or treatment process influent and effluent BOD is  
standard practice to measure its performance. Typical domestic wastewater BOD is  
about 200-250 mg/l.  
The oxygen consumed by the microorganisms during the BOD test is primarily for:  
1) Oxidizing the carbonaceous material (cBOD – carbonaceous BOD), and 2) Oxidiz-  
ing nitrogenous constituents such as ammonia (nBOD – nitrogenous BOD).  
Thus, BOD (Total) = cBOD + nBOD. The cBOD and nBOD is measured by adding  
certain chemical inhibitors which will inhibit the bacteria responsible for consuming  
the nitrogenous matter, thus measuring only the cBOD as part of the BOD test.  
Since not all of the organics is metabolized in the 5 days of the regular BOD test,  
certain wastewater discharge permits require reporting of the ultimate BOD value  
(BODU )  
Chemical Oxygen Demand (COD)  
The COD test involves using chemical oxidizers to measure the oxygen demand of  
the wastewater.  
As the chemical oxidizers will oxidize other constituents present, including inorganic  
matter, the COD value of wastewater will be higher than the BOD.  
The COD test can be conducted rather quickly than the 5 day BOD test, it is an effec-  
tive method to quantify the wastewater strength and process efficiencies and allow  
operators to make timely process adjustments.  
Total Organic Carbon (TOC)  
The TOC method utilizes laboratory analytical instruments which directly measure the  
organic carbon content by quantifying the amount of carbon dioxide produced from the  
complete combustion of the organics present.  
Note: BOD measures the amount of oxygen required by the microorganisms present to consume  
the organic material while COD measures the chemical oxidation required to oxidize all chemicals  
including organics present in wastewater. BOD value of typical domestic sewage is about 200 -  
250 mg/l while the COD value ranges from 300 - 450 mg/l. Typical BOD:COD ratio ranges from  
0.5-0.8.  
4.1 Wastewater Constituents  
4.1.2 Solids  
29  
Like BOD, wastewater solids is another critical parameter for establishing the wastewater strength  
and determining treatment process efficiencies.  
The solids can be classified as suspended or dissolved based upon its ability to pass through  
a standardized filter paper.  
• When the wastewater is filtered:  
the residual solids remaining on the filter paper after drying in an oven at 103C is  
the suspended solids portion, and  
the solids remaining after drying the filtrate are the dissolved solids.  
Suspended solids include larger floating particles and consist of sand, grit, clay, fecal  
matter, paper, pieces of wood, particles of food and garbage, and similar materials.  
• Suspended solids can be categorized based upon its settling characteristics as:  
Settleable  
Non-settleable  
Colloidial-small, charged (typically negative) particles which do not settle easily.  
Some of the colloidial particles are small enough to pass through the filter paper  
used for filtering the suspended solids  
*
Floatable-example oil and grease and small plastics  
*
Dissolved solids in wastewater include organics. However, the major elements of dissolved  
solids are inorganic ions such as Ca+2, Mg+2, Cl, SO4 2 , HCO3 , Fe+2, PO4 3, NO3  
. These ions are part of the dissolved salts such as sodium chloride (NaCl), calcium bi-  
carbonate (Ca(HCO3)2), magnesium phosphate (Mg3PO4) and others which are normally  
present in water and wastewater.  
Conductivity or electrical conductance (EC) measurement is typically conducted as  
the wastewater enters the plant as conductivity provides an indirect and simple measure  
of the amount of dissolved solids present.  
Conductivity or electrical conductance (EC) is a measure the amount of electrical  
current that can be conducted by a solution.  
The conductance of electricity in a solution is due to the presence of dissolved inor-  
ganic ions  
The higher the concentration of these ions, the higher is the conductivity.  
Conductivity is measured in the units of mhos/cm or Siemens/cm. (Note: mhos is the  
reverse of ohm which is a measure of resistance).  
Typical wastewater conductivities range from 50 to 1500 S/cm  
• Both suspended and dissolved solids can be either volatile (organic) or fixed (inorganic).  
• Total Solids is thus a sum of TSS and dissolved solids or volatile and fixed solids.  
The volatile solids are typically of plant or animal origin .  
The fixed solids include sand, gravel and silt as well as the dissolved salts.  
 
30  
Chapter 4. Constituents Properties and Analysis  
The volatile or fixed fractions are quantified  
by incinerating the solids in a muffler furnace  
at 550which removes only the volatile solids  
leaving only the fixed solids behind.  
In terms of the size of the solids, the distribu-  
tion is approximately thirty percent suspended  
and about seventy percent dissolved solids -  
which includes the colloidal particles which  
have passed through the filter paper.  
As primary treatment process involve settling of  
solids, establishing the settleable portion of the  
suspended solids is important.  
Imhoff Cone  
The settleable solids are quantified using an  
Imhoff cone and are reported in ml/L. Imhoff  
cone is a 1 liter, clear cone shaped container,  
with volume graduations (ml) at the bottom.  
One factor which affects settleability is the conveyance time of the sewage to the treatment  
plant.  
Note the ml markings at the bottom of the cone  
The settleable component of the suspended solids will decrease as the sewage becomes  
more septic due to longer conveyance times.  
Influent and effluent total suspended solids are measured to establish the overall treatment  
and individual process efficiencies.  
Volatile solids measurements before and after biological processes such as secondary  
treatment and digestion provide information on the process efficiency.  
Summary of Wastewater Solids  
• Solids in wastewater can be categorized as dissolved or suspended  
Suspended solids can be further categorized as settleable or unsettleable  
• Solids can also be categorized as organic (aka: volatile) or inorganic (aka: fixed)  
Colloidial particles are small sized particles some of which pass through the filter and  
accounted as part of dissolved solids  
TSS - Total Suspended Solids are the solids that are captured on the filter paper upon  
filtration of the wastewater sample.  
Wastewater samples typically analyzed for TSS include: plant, primary and secondary  
processes - influent and effluent. TSS is reported in mg/l  
TS - Total Solids are solids content of sludge. TS of sludge is established by drying a  
preweighed quantity of sludge in an oven and is typically reported as % solids - which is  
how many parts (by weight) of solids per 100 parts (by weight) of sludge.  
Volatile solids are solids that are removed when the solids are incinerated at 550C. The  
solids that remain after incineration are fixed or non-volatile or inorganic solids.  
4.1 Wastewater Constituents  
31  
Wastewater Solids Content  
Less than 0.1% total solids. Total solids concentration in typical wastewater is about  
750mg/l  
• The total solids are 50% organic (volatile) and 50% inorganic (fixed)  
Of the total solids, dissolved solids constitute about 70% of the solids and the remaining  
30% solids are suspended solids  
• 40% of the dissolved solids are volatile the remaining 60% are fixed  
• 70% of the suspended solids are volatile and the remaining 30% are fixed  
Figure 4.1: Typical Wastewater Solids Concentrations  
4.1.3 Nutrients  
Plant nutrients - nitrogen and phosphorous, present in wastewater effluent discharge, pro-  
mote growth of plant and algal matter in the receiving waters causing destruction of the  
normal aquatic life mainly due to oxygen depletion - eutrophication.  
Because of the potential impacts of the presence of these nutrients in wastewater effluent  
on the receiving waters, limits on the levels of these nutrients is typically stipulated in the  
treatment plant’s wastewater discharge permit.  
Typically, conventional secondary treatment processes are designed primarily remove  
the organics from the wastewater. Secondary treatment process designed to additionally  
remove nutrients is deemed as tertiary or advanced treatment is termed as Biological Nutri-  
ent Removal (BNR).  
 
32  
Chapter 4. Constituents Properties and Analysis  
Nitrogen  
Forms of nitrogen:  
About 60% of nitrogen in wastewater is present as ammonia nitrogen (about 60%). The  
ammonium nitrogen is present either in the form of ammonia (NH3 ) or as ammonium  
(NH+4 ) ion. These two forms can rapidly change from one to the other depending on pH  
and temperature. Under low pH (acidic) or neutral conditions – pH less than or equal to 7,  
ammonia exists mostly as ammonium. Ammonia becomes the dominant form as the pH  
increases to 8 and beyond.  
The other dominant form of nitrogen, about 40% of the total nitrogen is as organic nitro-  
gen  
Nitrogen measured as Total Kjeldahl Nitrogen (TKN) which is the sum of the organic  
nitrogen and the ammonia nitrogen concentrations. Total inorganic nitrogen is the total  
concentration of ammonia nitrogen, NO3-, and NO2-. Table provides the concentrations  
and forms of nitrogen in wastewater.  
Forms of Nitrogen in Wastewater  
Forms of Nitrogen  
Formula  
Found in  
Typical  
Concentra-  
tion  
30-50 mg/l  
+
Ammonia/Ammonium  
NH3/NH4  
Influent wastewa-  
ter  
Wastewater  
effluent  
Total Kjeldahl Nitrogen  
(Ammonia/Ammonium + Organic Nitro-  
TKN  
30-60 mg/l  
gen)  
Total Inorganic Nitrogen  
TIN  
Wastewater  
effluent  
1-40 mg/l  
(Ammonia/Ammonium + Nitrite + Ni-  
trate)  
Nitrate  
Nitrate  
NO3 −  
Nitrified effluent  
Partially nitrified  
effluent  
1-35 mg/l  
0.1-2 mg/l  
NO2  
Figure 4.2: Forms of Nitrogen  
Phosphorous  
Forms of phosphorous:  
The principal forms are organically bound phosphorus, polyphosphates, and orthophos-  
phates.  
Organically bound phosphorus originates from body and food waste and, upon biological  
decomposition of these solids, is converted to orthophosphates.  
Polyphosphates originate from synthetic detergents and are hydrolyzed to orthophosphates.  
Thus, the principal form of phosphorus in wastewater is assumed to be orthophosphates,  
3−  
although the other forms may exist. Orthophosphates consist of the negative ions PO4  
,
4.2 Wastewater Properties and Parameters  
33  
HPO42, and H2PO4 . These may form chemical combinations with cations (positively  
charged ions).  
Oil and Grease  
Fats, oil and grease in wastewater originate from homes, food establishments and industries.  
Oil and grease content of wastewater is established in the laboratory by extracting it with a  
solvent - n-hexane. The concentration of oil and grease is reported in mg/l and typical oil  
and grease content of wastewater ranges from 80 - 120 mg/l  
Presence of excessive oils and grease could potentially impact the secondary treatment  
process  
Oils and grease are removed as floatables in primary treatment and sent with the sludge to  
the digesters  
4.2 Wastewater Properties and Parameters  
Laboratory and field tests are conducted to measure parameters which are critical for monitoring  
and controlling treatment. The following are the key parameters that are measured.  
4.2.1 pH  
pH is a measure of the hydrogen ion (H+) content or the acidity or basicity of a solution. pH  
impacts the chemical and micribiological elements of wastewater treatment processes and thus  
pH measurement and control is critical.  
• Pure water dissociates into equal concentration of hydrogen ions and hydroxide ions:  
H2O H+ +OH.  
• The H+ are responsible for acidic properties and the OHions for the basic properties.  
pH is the inverse of H+ concentration; pH increases when the concentration of H+ de-  
creases relative to the concentration of OH-.  
pH scale ranges from 0 – 14. When the concentration of both H+ and OHare equal, as in  
pure water, it is considered neutral and its pH is 7.0.  
If the pH of a sample solution is below 7.0, the sample is termed acidic and is alkaline or  
basic if its pH is above 7.0.  
Each change of 1 pH unit represents a 10 fold change in concentration. For example, a  
sample with a pH of 2.0 is 1000 times more acidic than a sample with a pH of 5.0.  
pH is measured by an electrode that is sensitive only to H+ or using a pH strip which is  
essentially an adsorbent paper which is pre-impregnated with chemicals which change  
color under different H+ concentrations.  
Most organisms involved in biological wastewater treatment processes do well within a a  
narrow range of pH near neutral (pH of 7).  
 
 
34  
Chapter 4. Constituents Properties and Analysis  
4.2.2 Oxidation Reduction Potential (ORP)  
ORP measurements are common in wastewater process control for monitoring conditions  
and process efficiency  
• ORP is measured in milivolts (mV) using a probe  
ORP is a measure of the potential of oxidation/reduction – electron transfer, based chemi-  
cal reactions to occur  
If the measured ORP value (in mV), is positive it indicates an environment where oxida-  
tion will occur and if negative, an environment where reduction reactions will occur  
Higher positive value indicates a stronger oxidative environment and likewise, a lower  
(more negative) ORP value indicates a stronger reductive environment  
For example, during chlorine disinfection, which is an oxidation process, the wastewater  
will exhibit a positive ORP. Stronger the oxidative power of chlorine, higher will be the  
wastewater ORP value  
All living matter, including microbes depend upon respiration to generate energy and  
respiration involves series of chemical oxidation-reduction reactions  
Bacteria grow and thrive only in specific chemical - oxidative-reductive environments  
which support its inbuilt metabolic pathways  
Aerobic bacteria need molecular oxygen as the terminal electron acceptor as part of its  
cellular respiration proces. Bacteria adapted to exist in an environment where molecular  
oxygen is not present (anoxic and anaerobic), rely on electron acceptors such as NO3 (den-  
itrification), SO4 (sulfide formation) and carbon (methane formers in anaerobic digestion)  
Aerobic bacteria responsible for cBOD removal in the secondary treatment process would  
be inhibited or wiped-out if the wastewater oxidation potential dropped and become re-  
ductive. Likewise, if the wastewater in the sewer pipes which is normally of reductive  
(negative ORP) was to become oxidative because of aeration (dissolving oxygen) it would  
cease the hydrogen slufide activity of the anaerobic bacteria  
Typical Wastewater Process ORPs  
Clorine disinfection  
+650 to +700 mV  
+100 to +350 mV  
+20 to +250mV  
+50 to +250 mV  
+50 to -50 mV  
- 200 mV  
Nitrification  
Biological phosphorous removal  
Activated sludge cBOD degradation with free molecular oxygen  
Denitrification  
Influent wastewater  
Sulfide formation  
-50 to -250 mV  
-100 to -225 mV  
-100 to -250 mV  
-75 to -400 mV  
Anaerobic Digestion: Acid formation (Acidogenesis)  
Biological phosphorous removal  
Anaerobic Digestion: Methane production (Methanogenesis)  
 
4.2 Wastewater Properties and Parameters  
35  
4.2.3 Chlorine Residual  
• Chlorine in one form or another is the principal disinfecting agent employed.  
An important additional advantage over some other disinfectants is that chlorine leaves a  
disinfectant residual that assists in preventing recontamination during distribution, trans-  
port, and household storage of water.  
The absence of a chlorine residual in the distribution system may, in certain circumstances,  
indicate the possibility of post-treatment contamination.  
• Three types of chlorine residual may be measured:  
1. Free chlorine - the most reactive form, which is the chlorine present as hypochlorite  
(OCl), hypochlorous (HOCl) or a combination of the two.  
2. Combined chlorine - the less reactive but more persistent form, consisting of chlo-  
rine that is combined with ammonia, nitrogen, or nitrogenous compounds (chlo-  
ramines). This is the amount of chlorine that has reacted with nitrates and is unavail-  
able for disinfection.  
3. Total chlorine - which is the sum of the free and combined chlorine residuals.  
Free chlorine is unstable in aqueous solution, and the chlorine content of water samples  
may decrease rapidly, particularly at warm temperatures. Also, exposure to strong light or  
agitation will accelerate the rate of loss of free chlorine. Water samples should therefore be  
analyzed for free chlorine immediately on sampling and not stored for later testing.  
• DPD - N,N-diethyl-p-phenylenediamine, is a commonly used method of measuring the  
chlorine residual in water. DPD reacts directly with disinfectants (e.g. chlorine, chlo-  
ramines etc.) to produce a pink colored solution. The intensity of this colored solution is  
proportional to the concentration of disinfectant in the sample.  
• Amperometric sensors are also used for chlorine measurements.  
• Advantages of amperometric method include:  
Chemical reagents are not required  
Sensors are relatively free from interference from color, turbidity, and interference  
from iron, manganese, nitrate and chromates present in the sample.  
4.2.4 Alkalinity  
• Alkalinity is the ability of a water to neutralize acids.  
During certain wastewater treatment processes including anaerobic digestion, acids are  
generated as a result of microbiological activity. The bacteria and other biological entities  
which play an active role in wastewater treatment are most effective at a neutral to slightly  
alkaline pH of 7 to 8. In order to maintain these optimal pH conditions for biological  
activity there must be sufficient alkalinity present in the wastewater to neutralize acids  
generated by the active biomass.  
This ability to maintain the proper pH in the wastewater as it undergoes treatment is the  
reason why alkalinity is so important to the wastewater industry.  
 
 
36  
Chapter 4. Constituents Properties and Analysis  
The alkalinity is due to the presence of acid neutralizing bases in the water including the  
hydroxyl (OH), carbonate (CO3) and bicarbonate (HCO3) ions. These ions are of  
mineral origin and are also formed from carbon dioxide which comes from the atmosphere  
and from the microbial decomposition of organic material. The resistance to pH change of  
the water will continue until all the alkalinity contributing ions are neutralized.  
The pH of a water serves as a guide to the types of alkalinity present in the water but is  
unrelated to the alkalinity content of a water. Important Note: Alkalinity is a measure of  
the ability to neutralize acids whereas a solution is termed alkaline (or basic) if its pH  
greater than 7.  
• Alkalinity is expressed as milligrams per liter of CaCO3  
4.2.5 Dissolved Oxygen  
Dissolved oxygen (DO) is the concentration of oxygen dissolved in the wastewater sample  
and is typically measured in the field using an amperometric DO probe which utilizes an  
oxygen permeable membrane that enables a chemical reduction reaction, which produces  
an electrical signal to capture the DO concentration value.  
The presence of oxygen indicates an aerobic environment where dissolved, free oxygen is  
available for aerobic microorganisms to live, BOD removal in the activated sludge process  
occurs as a result of the activity of aerobic bacteria. The absence of DO indicates that the  
environment or condition is either anoxic or anaerobic.  
In an anoxic environment, free oxygen is not present, but oxygen is available from  
its combined forms - nitrate (NO3 ) and sulfate (SO4 ) for the the consumption of  
microorganisms. Example of an anoxic process is denitrification. In denitrification, the  
anoxic bacteria in the presence of food (cBOD) consume the combined oxygen in nitrates  
(NO3 ) and convert it to nitrogen gas.  
The complete absence of oxygen including free and combined oxygen is an anaerobic  
environment.  
Microorganisms are termed as obligate aerobes if they cannot survive without free oxygen.  
Facultative aerobes are microorganisms which can survive in both aerobic and anaerobic  
environments.  
4.2.6 Microbiological testing and monitoring  
Microbes play a critical role in wastewater treatment.  
Heterotrohic (organisms that consume organic material) microbes are responsible for the  
biological wastewater treatment processes - secondary treatment process, digestion and  
nutrient removal; and  
• Pathogens - agents that cause disease are present in wastewater effluent.  
Microbiological testing and monitoring is conducted as part of the wastewater treatment typically  
for the following:  
1. Microbiological testing related to monitoring and troubleshooting biological wastewater  
 
 
4.2 Wastewater Properties and Parameters  
37  
treatment  
Microbes involved in biological wastewater treatment processes include:  
Fungi - Filamentous fungi occasionally bloom in activated sludge processes due to  
low pH or nutrient deficiency and cause problems with the settleability.  
Protozoa - Protozoas play a important role in the secondary treatment process. Com-  
mon protozoas in the activated sludge process include:  
Amoeba  
Flagellate  
Cilliate  
• Rotifers  
• Nematodes  
Bacteria - Bacteria is the predominant microorganism responsible for the biological  
wastewater water treatment.  
The effectiveness of the biological wastewater treatment processes is primarily due to  
the presence of a microbial ecosystem with a right balance of populations of different  
microbial species.  
Methods used for monitoring the microbial composition include direct monitoring us-  
ing a light microscope to see which and how many of the different microbial species  
are present - typically used for activated sludge process.  
Indirect method includes monitoring other parameters such as pH and alkalinity  
which are influenced by microbiological activity.  
The microbial monitoring ensures process stability and helps identify potential pro-  
cess upset conditions caused by changes to the microbial population due to other  
external factors - toxicty, organic loading, temperature etc.  
2. Microbiological testing related to monitoring and controlling pathogens in treated wastew-  
ater effluent  
Pathogens in wastewater belong to the following groups:  
Bacteria: Although, bacteria is present in large numbers in feces, pathogenic or bac-  
teria are present only because of an infection and this pathogenic bacteria can poten-  
tially spread the infection to other healthy individuals. Disease spread by pathogenic  
bacteria include diarrhea, cholera and typhoid among many others.  
Viruses: A large number of viruses may infect humans and are present in feces.  
These include enteroviruses (including polioviruses), hepatitis A virus, reoviruses  
and diarrhea-causing viruses (especially rotavirus).  
Protozoa: Many species of protozoa can infect humans and cause diarrhoea and  
dysentery. Girardia which casues diarrheal illness is an example of a protozoan  
pathogen  
Helminths: These are parasitic worms that can infect humans and are transmitted to  
others through its eggs or larval forms  
As one of the main reasons for treating wastewater is to protect public health, micro-  
38  
Chapter 4. Constituents Properties and Analysis  
biological/pathogen testing of the wastewater effluent and the surface water impacted  
by the wastewater discharge is conducted to meet the requirements of a wastewater  
discharge permit, to monitor the pathogen impact of treated wastewater discharge and  
assess the level of contamination of a public body of water.  
The bacteriological tests involves detection and quantification of one or more of the  
following bacteria: total coliforms, fecal coliforms, E. Coli, and Enterococcus.  
The main reason why these bacteria such as coliforms and enterococcus are used  
as it is not practical to detect and quantify all pathogens associated with wastewater.  
These selected bacteria originate from feces and indicate fecal contamination  
and thus serve as an indicator organisms for pathogens of wastewater origin.  
Also, they are abundant, potentially less harmful, and easy to detect. E. coli has  
been shown to be a better predictor of the potential for impacts to human health  
and therefore many newer wastewater discharge permits require E. Coli testing  
in lieu of fecal coliform testing requirements.  
The microbiological test sample is always collected as a grab in a clean, sterile  
borosilicate glass or plastic bottle containing sodium thiosulfate.  
Sodium thiosulfate is added to remove residual chlorine which will kill col-  
iforms during transit.  
If the sample is not preserved or maintained under proper conditions until the  
test is conducted in the laboratory, the test would provide erroneous results.  
Samples must be refrigerated if they cannot be analyzed within 1 hour of col-  
lection and must be handled with care to prevent contamination and adverse  
conditions such as prolonged exposure to direct sunlight.  
The maximum holding time for state or federal permit reporting purposes is 6  
hours.  
As it is not possible to exactly quantify the number of bacteria present, a statistical  
based - Most Probable Number (MPN) approach is utilized. The methods for wastew-  
ater bacteriological tests include: multiple-tube fermentation technique, membrane  
filtration and quanti-tray testing.  
4.2.7 Specific Gravity  
• Specific gravity is a term to express the weight of a solution with respect to that of water  
• Water weighs 1 kg/L or 8.34 lbs/gallon or 62.4 lbs/ft3  
A solution with a specific gravity of 1.2 will weigh 1.2 times the same volume of water. 1  
L of that solution will weigh ( 1.2 kg )/L or ( 1.2*8.34=10lbs )/gallon.  
Typically wastewater and the associated unthickened sludge, for all practical purposes is  
assumed to have a specific gravity of 1 - implying 8.34 lbs/gallon.  
Specific gravity is typically used for calculations related to chemicals used in wastewater  
treatment.  
 
4.3 Wastewater Sampling  
39  
4.3 Wastewater Sampling  
Field or laboratory measurement of a certain parameter is critical in wastewater treatment  
operations to obtain information about wastewater characteristics in order to either charac-  
terize a wastewater stream, or to monitor a treatment process or for permit compliance.  
A sample is a small part of the whole representing the whole. Thus, a sample needs to be  
such that it truly represents the entire population – which in a wastewater operations could  
be either a wastewater stream, wastewater solids or a chemical used.  
4.3.1 Sampling Methods  
Grab Samples  
• A grab sample is a sample collected at a specific spot at a site over a short period of time.  
Grab sampling allows for instantaneous analysis of parameters such as pH, dissolved  
oxygen, chlorine residual, temperature and other parameters which change rapidly with  
time.  
• A grab sample represents a snapshot of space and time of a process stream.  
Composite Samples  
A composite sample is a collection of discrete samples are combined over a certain period  
or space and therefore represent the average performance of a wastewater treatment plant  
or a process during the collection period.  
• Composite sampling can be either based on:  
1. constant time interval (time proportioned sampling)  
2. constant wastewater volume interval (flow-proportioned sampling), and  
3. treatment process space - includes samples taken at different depths  
Composite samples are typically collected using automated samplers which can be pro-  
grammed to collect samples at pre-established time intervals – for time proportional sam-  
pling.  
Time and space composite samples are collected by adding equal volumes of samples  
collected from different times or locations.  
Flow proportional composite samples comprise of volume of each subsample based on  
flow.  
Grab Sampling Using a Long Handle Dipper  
Automated Sampler  
 
 
40  
Chapter 4. Constituents Properties and Analysis  
Sampling Precautions and Protocols  
Samples should represent the major portion of the process or the process stream and  
should be taken from places where the mixing is thorough, avoiding dead spots and ar-  
eas of heavier or lighter loadings.  
The collected sample is invariably exposed to conditions very different from the original  
source and is subject to change due to chemical and microbiological activity.  
Thus, in order to ensure integrity of sample, sample preservation techniques specific to the  
analysis to be performed is needed.  
The preservation technique should not only allow for stabilizing the parameter to be  
analyzed, it should also not interfere with the analyses.  
The common preservation techniques involve use of proper containers, temperature  
control, addition of chemical preservatives, and observance of the recommended  
maximum sample holding time.  
Bacteriological Sampling  
• Always collected as a grab  
A clean, sterile borosilicate glass or plastic bottle containing sodium thiosulfate is used.  
Sodium thiosulfate is added to remove residual chlorine which will kill coliforms during  
transit. If the sample is not preserved or maintained under proper conditions until the test  
is conducted in the laboratory, the test would provide erroneous results  
• Samples must be refrigerated if they cannot be analyzed within 1 hour of collection  
Samples must be handled with care to prevent contamination and adverse conditions such  
as prolonged exposure to direct sunlight  
• Maximum holding time for state or federal permit reporting purposes is 6 hours  
4.3.2 Data Reporting  
Arithmetic mean is typically calculated for reporting data where multiple samples have  
been collected and analyzed for the same process stream at different times and for report-  
ing average value over a certain time period – daily, monthly etc.  
Arithmetic mean mathematically is calculated by adding all the result values and dividing  
by the total number of data points.  
Mathematically the arithmetic mean is represented as:  
n
i=1 xi x1 +x2 +x3...xn  
x¯ =  
=
n
n
For example:  
Arithmetic mean of the following set of data points: 200, 304, 250, 400 is calculated as:  
200+302+250+400  
Arithmetic Mean =  
= 288  
4
For data sets for analysis such as fecal coliform could include values which vary by several  
 
4.4 Laboratory Analysis  
41  
orders of magnitudes, using the arithmetic mean to report the average value is not appropriate as  
the lower or higher values would bias the calculated mean.  
For example, consider a data set with values: 260, 300, 500, 5,000, 320 and 200.  
260+300+500+5,000+320+200  
The arithmetic mean =  
= 3,444  
6
Here the 5000 value completely skews the arithmetic mean.  
Therefore, for such tests, the geometric mean calculation is used for reporting the average value.  
Mathematically a geometric mean is represented as:  
1
n
 
!
n
=
n a1 a2 a3...an  
i=a  
Calculation method:  
1. Find the product of all the data points (analogous to first calculating the sum of all the data  
points when calculating the arithmetic mean)  
260*300*500*5,000*320*200 = 12,480,000,000,000,000  
2. Raise the product to the inverse of the number of data points  
(*Using the power function of a scientific calculator)  
1
6
Here n (# data points) = 6 =geometric mean = (12,480,000,000,000,000) = 482  
4.4 Laboratory Analysis  
4.4.1 BOD Analysis  
The Biochemical Oxygen Demand (BOD) test estimates the amount of biodegradable ma-  
terial present by measuring the amount of oxygen used by the bacteria to break down the  
organic waste in the sample incubated at 20 deg. C over a five-day period . The BOD test  
provides an indication on the strength of wastewater in terms of how much oxygen could  
be depleted if that wastewater was introduced into another receiving water. Complete sta-  
bilization of a sample may require a period of incubation too long for practical purposes;  
therefore, 5 days has been accepted as the standard incubation period.  
As the regular BOD test includes estimation of oxygen nitrifying bacteria consumes in the  
process of converting inorganic forms of ammonia and nitrogen to nitrite and nitrate, its  
value represents oxygen used for removing both, organic material and nitrogenous mat-  
ter. As this BOD value does not quite represent the organic strength of the wastewater,  
the normal BOD test is modified by introducing a chemical inhibitor - 3 mg of 2-chloro-  
6-(trichloro methyl) pyridine (TCMP), which suppresses the growth of the nitrogenous  
bacteria so that the resultant BOD measured represents the oxygen depletion associated  
with the depletion of the organic matter only. This is the Carbonaceous biochemical oxy-  
gen demand or cBOD.  
 
 
42  
Chapter 4. Constituents Properties and Analysis  
Thus tBOD = nBOD + cBOD  
Wastewater BOD measurement involves testing a sample set consisting of several sample  
dilutions along with a "Blank". "Blank" is a sample with only the dilution water with no  
wastewater added.  
The dilutions are made based upon the expected BOD concentration of the sample. Using  
the final dilution volume of 300 ml, the initial sample volume can be estimated using the  
formula:  
h
ꢁi  
mg  
Oxygen Depletion  
l
Sample Volume(ml) =  
300 ml  
mg  
Anticipated BOD  
l
For example, if testing an influent wastewater BOD with an expected BOD value of  
mg  
4
l
250 mg/l, a range of sample volumes for dilution around sample volume of  
mg  
l
250  
300 ml=5ml.  
The data obtained for each of the dilutions after the 5-day incubation period must meet the  
following criteria for the sample value to be acceptable for calculating the BOD.  
1. A residual DO of at least 1 mg/L,  
2. A DO depletion of at least 2 mg/L  
Additionally, the whole sample set is rejected if the Blank shows an oxygen depletion of  
>0.2mg/l.  
• BOD is calculated for each sample dilution value using the following formula:  
mg  
l
Initial DODO Day 5  
300 ml  
Sample Volume (ml)  
BOD  
=
4.4.2 Wastewater solids  
Total (TSS) and Volatile (VSS)  
• A known volume of wastewater sample is filtered through a pre-weighed filter paper  
• The suspended solids will be retained by the filter  
• The water with the dissolved solids will pass through the filter  
• The filter paper with the filter solids is rinsed with distilled water to remove  
 
4.4 Laboratory Analysis  
43  
• The filter paper with the solids is dried in the oven and then weighed  
The difference between the weight of the dried filter paper with the solids and the pre-  
weighed filter paper, measured in mg, will be the suspended solids in: mg per the original  
quantity of wastewater sample taken. This value can be converted to give the suspended  
solids content in mg/l  
• A filter paper with the dried solids is incinerated in a muffler furnace  
• The difference in the weight of the solids, before and after incineration is the fixed solids  
The difference between the weight of the solids before incineration and the fixed solids is  
the volatile solids  
¨
weight of solids gms 1000 ml 1000 mg  
¨
mg  
l
Total Suspended Solids - TSS TSS  
=
¨
gms  
¨
volume of sample ml  
l
weight of filter paper with dried solidsweight of filter paper  
volume of sample (ml)  
=
1,000,000  
Volatile Suspended Solids - VSS  
mg weight of volatile solids gms 1000 ml 1000 mg  
¨
¨
VSS  
=
¨
gms  
¨
l
volume of sample ml  
l
wt. of filter paper with dried solidswt. of filter paper incinerated residue  
volume of sample (ml)  
=
1,000,000  
(
(
(
(
(
(
weight (gms) of volatile solids gms volatile solids 100 gms total solids  
(
(
(
VSS(%) =  
=
(
(
(
(
(
(
100 gms total solids  
100 gms total solids  
gms total solids  
(
(
wt. o f filter paper with dried solidswt. of filter paper incinerated residue  
=
100  
wt. o f filter paper with dried solidswt. of filter paper  
Wastewater and Sludge Total & Volatile Solids  
A certain quantity of wastewater (by volume) or sludge (by weight) is taken in a pre-weighed dish  
and weighed. Note: the sample is not filtered.  
• The dish with the sample is dried in an oven  
The difference in the weight of the pre-weighed dish from that of the dish with the dried sample is  
the total solids  
• The dried solids are incinerated in a muffler furnace  
• The difference in the weight of the solids, before and after incineration is the fixed solids  
• The difference between the fixed solids and the total solids is the volatile solids  
Total solids of a sludge sample is reported as a % of the sludge weight. A 7% sludge has 7 lbs of  
44  
Chapter 4. Constituents Properties and Analysis  
solids for every 100 lbs of sludge.  
For sludge samples, volatile solids is typically reported as the volatile solids fraction in % of the total solids  
content of the sludge. For example, if a 8% sludge (i.e sludge which has 8% TS or 80,000mg/l solids), is  
reported to have 70% volatile, it means that 70% of the total solids - 0.7*8%=5.6% or 56,000mg/l is the  
sludge volatile solids content. 70% volatile does not meet the sludge has 700,000mg/l volatile solids  
weight o f solids (gms)  
100 gms o f sample  
gms solids  
Total Solids - TS TS(%) =  
=
100  
gms sample  
weight o f cruicible with dried solidsweight ofcruicible  
weight o f cruicible with sampleweight ofcruicible  
=
100  
weight of volatile solids (gms) gms volatile solids  
Total Volatile Solids - VS VS(%) =  
=
100  
gms total solids  
100 gms of total solids  
wt. o f cruicible with dried solidswt. of cruicible incinerated residue  
wt. o f cruicible with dried solidswt. of cruicible  
=
100  
4.4 Laboratory Analysis  
45  
Sample BOD and solids analysis math problems  
1. BOD tests are run on the final effluent from an activated sludge plant with and without the use of  
a "nitrification inhibitor". Three hundred milliliter bottles (300 ml) are used in these tests. The  
raw data for these tests are presented below. What percentage of the average total BOD is the  
average nBOD?  
Sample Volume, ml  
10  
20  
30  
40  
Blank  
Initial DO, mg/l  
Final DO, mg/l  
9.0  
6.9  
8.9  
4.8  
8.8  
2.5  
9.1  
1.1  
9.1  
9.0  
BOD Test with "inhibitor" added (cBOD)  
Sample Volume, ml  
10  
20  
30  
40  
Blank  
Initial DO, mg/l  
Final DO, mg/l  
8.9  
7.5  
8.9  
6.2  
9.0  
5.0  
9.0  
3.3  
9.1  
9.0  
Solution:  
Blanks for both tBOD and cBOD are both <=0.2mg/l - thus sample sets are acceptable  
Sample Volume, ml  
10  
20  
30  
40  
tBOD Diff., mg/l  
tBOD, mg/l  
2.1  
4.1  
6.3  
8
2.1*300/10  
=63.0  
4.1*300/20  
= 61.5  
6.3*300/30  
= 63.0  
8.0*300/40  
= 60.0  
cBOD Diff., mg/l  
cBOD, mg/l  
1.4  
2.7  
4.0  
5.7  
Reject  
2.7*300/20  
= 40.5  
4.0*300/30  
= 40  
5.7*300/40  
= 42.75  
Depletion < 2  
tBOD(avg) = (63+61.5+63+60)/4 = 61.9  
cBOD(avg) = (40.5+40+42.75)/3 = 41.1  
nBOD = tBOD - cBOD =nBOD = 61.9-41.1=20.8 =nBOD(%)=20.8/61.9*100= 33.6%  
46  
Chapter 4. Constituents Properties and Analysis  
2. Calculate percent total solids and percent volatile solids of a sludge sample given the following  
data:  
Weight of dish  
=
=
=
=
104.55 gms  
199.95 gms  
108.34 gms  
106.37 gms  
Weight of dish and wet sludge  
Weight of dish and dry sludge  
Weight of dish and ash  
Solution:  
Weight of dish=104.55 gms  
Weight of dish and wet sludge=199.95 gms  
Weight of dish and ash = 106.37 gms  
=Weight o f sludge = 199.95104.55 = 95.40 gms  
=Weight o f dry sludge (solids) = 108.34104.55 = 3.79 gms  
=Weight o f volatile solids = 108.34106.37 = 1.97 gms  
(
(
(
(
3.79 gms solids 100 gms sludge  
(
(
gms solids  
100 gms sludge 95.40 gm(s sludge 100 gms sludge  
Total solids(TS%) =  
=
= 3.97%  
= 52.0%  
(
(
(
(
(
(
(
(
(
1.97 gms volatile solids 100 gms total solids  
(
(
(
Total volatile solids(VS%) =  
(
(
(
(
(
3.79  
100 gms total solids  
gms total solids  
(
(
(
4.4.3 Bacteriological Enumeration  
Involves bacteriological testing of the wastewater effluent and the surface water impacted by the  
wastewater discharge  
• Conducted in-order to:  
1. Meet the requirements of a wastewater discharge permit  
2. Monitor the pathogen impact of treated wastewater discharge  
3. Assess the level of contamination of a public body of water  
4. Bacteriological tests involves detection and quantification of one or more of the following  
bacteria: total coliforms, fecal coliforms, E. Coli, and Enterococci.  
Wastewater Bacteria  
5. In wastewater, fecal coliforms originate in the intestines of warm-blooded animals. Aerobic  
bacteria including coliforms partake in the metabolization of the organic matter as part of the  
secondary treatment process  
6. Fecal coliforms are seldom pathogenic under normal circumstances and are easily cultured,  
their presence indicates the potential presence of pathogens  
 
4.4 Laboratory Analysis  
47  
The reason why these bacteria such as coliforms and enterococcus are used:  
(a) It is not practical to detect and quantify all pathogens associated with wastewater  
(b) These bacteria originate from feces and indicate fecal contamination and thus serve as  
an indicator organisms for pathogens of wastewater origin  
(c) They are also:  
abundant  
potentially less harmful, and  
easy to detect  
(d) E. coli has been shown to be a better predictor of the potential for impacts to human  
health and therefore many newer wastewater discharge permits require E. Coli testing in  
lieu of fecal coliform testing requirements.  
Bacteriological Testing Methods  
The methods for wastewater bacteriological tests include: multiple-tube fermentation (MTF) technique,  
membrane filtration (MF) and quanti-tray testing. When using the MTF and MF methods, it is not possi-  
ble to exactly quantify the number of bacteria present, a statistical based - Most Probable Number (MPN)  
approach is utilized  
The Multiple-Tube Fermentation (MTF) technique  
This involves adding three volumes – 10 ml, 1 ml and 0.1 ml of the sample, each to a set of five tubes  
containing Lauryl Tryptose broth and an inverted tube (Durham tube), followed by incubating the tubes at  
for a specified time. The Lauryl Tryptose broth produces color and/or turbidity change due to the growth  
of the target bacteria and the inverted tube collects the gas produced by the bacterial respiration. At the  
end of the process, the number of tubes showing bacterial growth are counted for each volume of sample  
and using this information the concentrations of organisms in the original sample are established using  
Statistical Tables. The test is conducted in three parts – presumptive, confirmative and completed. A  
schematic of the MTF used for quantifying total coliforms and fecal coliforms is provided below.  
Wastewater Sample  
Make required serial dilutions (See Note 1)  
Example: Expected coliform value range 5000 – 50000.  
First dilution (1:10): 11 ml sample + 99 ml dilution water  
Second dilution (1:100): 11 ml of first dilution + 99 ml dilution water  
Third dilution (1:1000): 11 ml of second dilution water + 99 ml dilution water  
Setup 15 tubes with Lauryl Tryptose Broth and an inverted vial (Durham tube)  
5 tubes each for 10 ml, 1 ml and 0.1ml sample  
10 ml  
1 ml  
0.1 ml  
Incubate 24 hrs – check tubes for positive results marked by : 1) broth turning cloudy, and 2) gas collection in  
the inverted tube. If no gas collection or media clouding is observed, incubate for another 24 hrs and then  
check again to see if the tubes are positive  
- + - - -  
- + - - -  
+ - - + +  
10 ml  
1 ml  
0.1 ml  
Inoculate each positive into bacteria specific broths – EC broth for fecal coliforms and Brilliant Green Bile (BGB)  
broth for total coliforms with inverted tubes. Incubate for upto 24 hrs for fecal coliforms and 24 hours for total  
coliforms and observe for positive results  
EC Broth (for fecal coliform}  
Note: EC Broth with MUG is used for E. Coli  
BGB Broth (for total coliform)  
Count number of confirmed positives for each set of the three sample volumes  
Example above: 10 ml – 3, 1 ml – 1, 0.1 ml - 1  
Establish the MPN using the statistical MPN Table  
Example 3-1-1 From Table - MPN = 14 MPN/100 ml. Given 1:1000 dilution, MPN = 14 * 1000 = 14,000 MPN/100 ml  
If after 24 hours, from the gram staining of the colonies  
Use colonies to innoculate an  
agar slant and nutrient broth  
from the agar slant, nonsporing gram negative bacteria  
can be identified and the lactose broth shows gas for-  
mation, the completed test is deemed positive  
Streak agar plates & incubate  
50  
Chapter 4. Constituents Properties and Analysis  
The Membrane Filtration (MF) method  
This is a faster way to estimate bacterial populations in water. In this method, an appropriate sample  
volume is passed through a membrane filter with a pore size small enough (0.45 micron) to retain the  
bacteria present. The filter is placed on an absorbent pad (in a petri dish) saturated with a culture medium  
that is selective for coliform growth. The petri dish containing the filter and pad is incubated, upside down,  
for 24 hours at the appropriate temperature. After incubation, the colonies that have grown are identified  
and counted using a low power microscope. A MUG medium is used for E- Coli. If E. Coli is present, it  
will make the MUG fluorescent when viewed in UV light.  
4.4 Laboratory Analysis  
51  
Quanti-trays tests  
This test used for the detection and quantification of specific microorganisms is being used increasingly  
mainly because it is a quicker test than the MTF. Colilert and Enterolert are the quanti tray based tests for  
E. Coli and Enterococcus. This method involve the use of specific enzymes and overcomes the drawbacks  
of the MTF which include false positives and negatives due to the more generic nature of the media used  
Chapter Assessment  
1. Positive ORP indicates the presence of an  
environment  
2.  
is the predominant microorganism responsible for biological wastewater treatment pro-  
cesses  
3. MPN is a statistics based method to estimate the concentration of viable bacteria is wastewater and  
it stands for  
4. Describe the three types of composite sampling and explain in your own words on how the sam-  
pling would be conducted for each of those methods  
5. Alkalinity helps in  
activity  
control when organic acids are formed as a result of microbiological  
6. Nitrogen in wastewater is typically present as  
and  
7. Volatile solids is determined by incinerating the solids at  
deg. C in a  
8. What would be the expected BOD concentration in the wastewater if per capita (person) generation  
is 0.15lb BOD per day per person and each person produces 80 gallons per day?  
(Hint: The answer is going to be in mg BOD/l. So multiply lb per day per person by the inverse of  
dallons per day - so that the unit is lbs per gallon and then convert that to mg/l)  
9. Describe the three types of composite sampling and explain in your own words on how the sam-  
pling would be conducted for each of those methods  
10. tBOD =  
BOD +  
BOD  
11. BOD stands for  
12. Alkalinity helps in Nderlinehspace1cm control when organic acids are formed as a result of micro-  
biological activity  
13. Nitrogen in wastewater is typically present as  
and  
14. Volatile solids is determined by incinerating the solids at  
ment]  
[temp] deg. C in a  
[equip-  
15. How does one estimate the wastewater sample size used for BOD testing  
16. Explain how is the nBOD quantified  
54  
Chapter 4. Constituents Properties and Analysis  
17. Explain why coliforms and enterococcus are used for wastewater bacteriological testing  
18. What is MPN and why is it used.  
19. Explain the steps involved in the MTF method  
20. What is the difference between volatile solids (VS) and volatile suspended solids (VSS)  
21. Solids that can be used as food by microorganisms  
22. Test method for determining total ammonium and organic nitrogen content:  
23. Inorganic solids are also called  
24. Oxygen in an anoxic environment is present as  
25. Conductivity is the measure of  
26. Negative ORP signifies  
solids  
environment  
27. Alkalinity helps in  
control when organic acids are formed as a result of microbiological activity  
28. An Imhoff cone is often used to measure the effectiveness of primary sedimentation.  
a. True  
*b. False  
29. At a primarily domestic wastewater treatment plant, the influent wastewater BOD is always greater  
than its chemical oxygen demand.  
a. True  
*b. False  
30. Dissolved oxygen in wastewater usually is referred to as combined oxygen.  
a. True  
*b. False  
31. Domestic wastewater generally contains only about 0.1  
*a. True  
b. False  
32. Grab or composite samples may be used interchangeable, whichever is most convenient and safest  
for all laboratory tests.  
a. True  
*b. False  
33. Wastewater with a pH of 12.0 to 14.0 would be too "acidic" for biological treatment.  
*a. True  
b. False  
34. Pre-aeration improves settleability in primary clarifier  
*a. True  
b. False  
35. Total solids are made up of dissolved and suspended solids both of which contain organic and  
inorganic matter  
*a. True  
b. False  
36. pH value less than 7 indicates an alkaline or basic condition  
a. True  
*b. False  
37. Conductivity is a very useful test for assessing sea water intrusion into sewer lines  
*a. True  
b. False  
38. Sample obtained for wastewater bacteriological testing is typically a composite sample  
4.4 Laboratory Analysis  
55  
a. True  
*b. False  
39. Samples collected for the analysis of COD, BOD, and pH should be acidified for preservation.  
a. True  
*b. False  
40. The MPN test is used to measure pathogen concentrations in wastewater.  
a. True  
*b. False  
41. The total solids in wastewater would be the combination of the fixed solids and the settleable solids.  
a. True  
*b. False  
42. Wastewater with a pH of 2.0 to 4.0 would be too "basic" for biological treatment.  
a. True  
*b. False  
43. Match the measurement units for each of the following:  
44. The MPN test measures the number of pathogens in a wastewater sample  
a. True  
*b. False  
45. The values of organic matter present in wastewater as measured by BOD and COD tests are typi-  
cally almost identical  
a. True  
*b. False  
46. In order to obtain valid results in coliform testing, the sample must be dechlorinated at the time of  
its collection.  
*a. True  
b. False  
47. A flow of 100 gallons per capita per day is often used for estimating flow into a wastewater treat-  
ment plant.  
*a. True  
b. False  
48. An Imhoff cone is used to measure settleable solids in units of mg/l.  
a. True  
*b. False  
49. A standardized method exists for the measurement of floatable solids.  
a. True  
*b. False  
50. In the test for coliform bacteria, samples are incubated for 5 days at 20 deg. C.  
a. True  
*b. False  
51. Coliform group organisms are found only in wastewater.  
a. True  
*b. False  
52. Where highly colored samples are involved, the determination of pH by use of a pH meter rather  
than by color-comparison is  
preferred.  
56  
Chapter 4. Constituents Properties and Analysis  
*a. True  
b. False  
53. What comes into a treatment plant must go out. This is the basis of the solids balance concept.  
*a. True  
b. False  
54. Receiving water measurements are used to determine the effect of the plant’s waste discharge on  
the receiving waters; therefore, it is necessary to measure both stream and plant  
effluent characteristics.  
*a. True  
b. False  
55. There are two types of samples that may be collected. One is called an integrated sample, and the  
other is referred to as a composite.  
a. True  
*b. False  
56. Coliform testing is typically conducted on a grab sample  
*a. True  
b. False  
57. Typical domestic wastewater BOD content is about 2000mg/l  
a. True  
*b. False  
58. BOD is a measure of the organic content of wastewater  
*a. True  
b. False  
59. ORP measurements can be used for controlling the disinfection process  
*a. True  
b. False  
60. MPN is used for enumerating wastewater bacteria and it stands for Measured Pathogen Number  
a. True  
*b. False  
61. For measuring dissolved oxygen in wastewater, it is advisable to use a composite sample  
*a. True  
b. False  
62. A sample to be used for pH measurements should be preserved by the addition of an acid prior to  
its analysis in the laboratory  
a. True  
*b. False  
63. BOD is a measure of the organic content of wastewater and it stands for Biological Oxidation  
Demand  
a. True  
*b. False  
64. Conductivity is measured in the units of millivolts using an electrochemical probe  
a. True  
*b. False  
65. Conductivity measurements provide an indirect way to measure total solids present in wastewater  
a. True  
4.4 Laboratory Analysis  
57  
*b. False  
66. Typical wastewater TSS concentrations are in the 2,000-2,500 mg/l range  
a. True  
*b. False  
67. A 24-hr flow proportioned sample may be collected for a fecal coliform test  
a. True  
*b. False  
68. Wastewater with a pH of 12.0 to 14.0 would be too "acidic" for biological treatment.  
*a. True  
b. False  
69. The laboratory measurement of volatile solids is a fair approximation of the organic content of the  
wastewater.  
*a. True  
b. False  
70. BOD5 and SS are both used to measure the strength of wastewater  
*a. True  
b. False  
71. Pre-aeration improves settleability in primary clarifier  
*a. True  
b. False  
72. pH value less than 7 indicates an alkaline or basic condition  
a. True  
*b. False  
58  
Chapter 4. Constituents Properties and Analysis  
73. A BOD test is run on a secondary effluent. The data for this test are given below. Assuming that the  
average BOD test result for an “inhibitor” on this same secondary effluent was 22 mg/L (cBOD),  
calculate the oxygen demand caused by nitrification. What percentage of the total BOD is the  
nitrogenous oxygen demand? On April 04, 2009 Exam  
SampleVolume,ml  
InitialDO,mg/l  
FinalDO,mg/l  
50  
9.0  
3.5  
75  
8.9  
1.2  
Blank  
9.2  
9.0  
• Solution:  
Dif ference,mg/l  
5.5  
5.5300/50 = 33  
7.7  
7.7300/75 = 30.8  
9.0  
tBOD,mg/l  
(33+30.8)  
Average:  
= 31.9mg  
l
2
tBOD = cBOD+nBOD =31.9 = 22 + nBOD =nBOD = 9.9  
%nBOD = 9.9/31.9100 = 31%  
4.4 Laboratory Analysis  
59  
74. BOD tests are run on the final effluent from an activated sludge plant with and without the use of a  
"nitrification inhibitor". Three hundred milliliter bottles (300 ml) are used in these tests. The raw  
data for these tests are presented below. What is the average NITROGENOUS BOD (NBOD)?  
Exam on April 04, 2009  
BOD Test without "inhibitor" (tBOD)  
SampleVolume,ml  
InitialDO,mg/l  
FinalDO,mg/l  
30  
9.0  
5.1  
60  
8.7  
1.2  
Blank  
9.1  
9.0  
BOD Test with "inhibitor" added (cBOD)  
SampleVolume,ml  
InitialDO,mg/l  
FinalDO,mg/l  
30  
9.0  
6.5  
60  
8.7  
3.5  
Blank  
9.1  
9.0  
• Solution:  
tBOD  
Dif f.,mg/l  
3.9  
7.5  
0.1  
0.1  
tBOD,mg/l  
cBOD  
3.9300/30 = 39  
7.5300/60 = 37.5  
Dif f.,mg/l  
tBOD,mg/l  
2.5  
5.2  
2.5300/30 = 25  
5.2300/60 = 26  
Average:  
(39+37.5)  
tBOD:  
cBOD:  
= 38.3mg  
2
l
(25+26)  
= 25.5mg  
2
l
mg  
l
tBOD = cBOD+nBOD =38.3 = 25.5 + nBOD =nBOD = 12.8  
60  
Chapter 4. Constituents Properties and Analysis  
75. Calculate the TSS of the secondary effluent given the following:  
Sample volume  
41 ml  
Tare weight of filter sample  
Filter + dried residue  
1.4604 gm  
1.4722 gm  
76. The technician quickly pours 41 ml of well mixed influent into the filter funnel. The tare weight of  
the filter is 1.4604 gm. After rinsing, drying, cooling, and weighing, the first dry weight is 1.4722  
gm. The filter is returned to the oven and dried, cooled, and weighed. The second dry weight is  
1.4700 gm. Calculate the TSS.  
• Solution  
mg  
l
(1.47221.4604)gm TSS  
mg  
gm  
1000ml  
1000  
= 288  
41ml  
l
4.4 Laboratory Analysis  
61  
77. BOD tests are run on the final effluent from an activated sludge plant with and without the use of  
a "nitrification inhibitor". Three hundred milliliter bottles (300 ml) are used in these tests. The  
raw data for these tests are presented below. What percentage of the average total BOD is the  
average nBOD? (10 points)  
SampleVolume,ml  
InitialDO,mg/l  
FinalDO,mg/l  
10  
9.0  
6.9  
20  
8.9  
4.8  
30  
8.8  
2.5  
40  
9.1  
1.1  
Blank  
9.1  
9.0  
BOD Test with "inhibitor" added (cBOD)  
SampleVolume,ml  
InitialDO,mg/l  
FinalDO,mg/l  
10  
8.9  
7.5  
20  
8.9  
6.2  
30  
9.0  
5.0  
40  
9.0  
3.3  
Blank  
9.1  
9.0  
Solution:  
tBODDif f.,mg/l  
tBOD,mg/l  
2.1  
4.1  
6.3  
8
2.1300/10 = 63.0 4.1300/20 = 61.5 6.3300/30 = 63.0 8.0300/40 = 60.0  
1.4 2.7 4.0 5.7  
4.0300/30 = 40 5.7300/40 = 42.8  
cBODDif f.,mg/l  
cBOD,mg/l [Discard dif f. < 2] 2.7300/20 = 40.5  
tBOD(avg) = 63+61.5+63+60 = 61.9  
cBOD(avg) = 40.5+40+42.75 = 41.1  
nBOD = tBOD - cBOD =nBOD = 61.9-41.1=20.8 =nBOD(%)=20.8/61.9*100= 33.6%  
62  
Chapter 4. Constituents Properties and Analysis  
78. BOD tests are run on the final effluent from an activated sludge plant with and without the use of a  
"nitrification inhibitor". Three hundred milliliter bottles (300 ml) are used in these tests. The raw  
data for these tests are presented below. What is the average NITROGENOUS BOD (NBOD)?  
Exam on April 04, 2009  
BOD Test without "inhibitor" (tBOD)  
SampleVolume,ml  
InitialDO,mg/l  
FinalDO,mg/l  
30  
9.0  
5.1  
60  
8.7  
1.2  
Blank  
9.1  
9.0  
BOD Test with "inhibitor" added (cBOD)  
SampleVolume,ml  
InitialDO,mg/l  
FinalDO,mg/l  
30  
9.0  
6.5  
60  
8.7  
3.5  
Blank  
9.1  
9.0  
Solution:  
tBOD  
Dif f.,mg/l  
3.9  
7.5  
0.1  
0.1  
tBOD,mg/l  
cBOD  
3.9300/30 = 39  
7.5300/60 = 37.5  
Dif f.,mg/l  
tBOD,mg/l  
2.5  
5.2  
2.5300/30 = 25  
5.2300/60 = 26  
Average:  
(39+37.5)  
tBOD:  
cBOD:  
= 38.3mg  
2
l
(25+26)  
= 25.5mg  
2
l
mg  
l
tBOD = cBOD+nBOD =38.3 = 25.5 + nBOD =nBOD = 12.8  
4.4 Laboratory Analysis  
63  
79. Calculate percent total solids and percent volatile solids of a sludge sample given the following  
data:  
Weight of dish  
=
=
=
=
104.55 gms  
199.95 gms  
108.34 gms  
106.37 gms  
Weight of dish and wet sludge  
Weight of dish and dry sludge  
Weight of dish and ash  
(Answer: TS = 3.97%. VS = 52%)  
Solution:  
Weight of dish=104.55 gms  
Weight of dish and wet sludge=199.95 gms  
Weight of dish and ash = 106.37 gms  
=Weight o f sludge = 199.95104.55 = 95.40 gms  
=Weight o f dry sludge (solids) = 108.34104.55 = 3.79 gms  
=Weight o f volatile solids = 108.34106.37 = 1.97 gms  
(
(
(
100 gms sludge  
(
gms solids  
gms solids  
(
3.79  
(
Total solids(TS%) =  
=
( ∗  
= 3.97%  
(
(
100 gms sludge  
95.40 gms sludge 100 gms sludge  
(
(
(
(
gms volatile solids 100 gms (total solids  
1.97  
(
(
(
Total volatile solids(VS%) =  
= 52.0%  
100 gms total solids  
(
(
3.79  
(
gms total solids  
(
(
64  
Chapter 4. Constituents Properties and Analysis  
80. What is percent volatile solids of a sludge sample given the following data:  
Weight of dish  
=
=
=
=
130.69 gms  
249.94 gms  
134.74 gms  
132.05 gms  
Weight of dish and wet sludge  
Weight of dish and dry sludge  
Weight of dish and ash  
Solution:  
Weight of dish=130.69 gms  
Weight of dish and wet sludge=249.94 gms  
Weight of dish and ash = 132.05 gms  
=Weight o f sludge = 249.94130.69 = 119.25 gms  
=Weight o f dry sludge (solids) = 134.74130.69 = 4.05 gms  
=Weight o f volatile solids = 134.74132.05 = 2.69 gms  
(
(
(
100 gms sludge  
(
gms solids  
gms solids  
(
4.05  
(
Total solids(TS%) =  
=
( ∗  
= 3.4%  
(
(
100 gms sludge  
119.25 gms sludge 100 gms sludge  
(
(
(
(
gms volatile solids 100 gms (total solids  
2.69  
(
(
(
Total volatile solids(VS%) =  
= 66.4%  
100 gms total solids  
(
(
4.05  
(
gms total solids  
(
(
4.4 Laboratory Analysis  
65  
81. Products that are non-biodegradable will have  
a. Same BOD  
as compared with biodegradable products  
*b. A lower BOD  
c. A higher BOD  
d. There is no relationship between BOD and biodegradability  
82. Coliform bacteria are  
a. Algae  
b. Coagulant aids  
*c. Indicators  
d. Sequestering agents  
83. What is percent volatile solids of a sludge sample given the following data:  
Weight of dish = 130.69 gms  
Weight of dish and wet sludge = 249.94 gms  
Weight of dish and dry sludge = 134.74 gms  
Weight of dish and ash = 132.05 gms  
*a. 38.3%  
b. 66.4%  
c. 42.6%  
d. 58.0%  
84. What is percent volatile solids of a sludge sample given the following data:  
Weight of dish = 130.69 gms  
Weight of dish and wet sludge = 249.94 gms  
Weight of dish and dry sludge = 134.74 gms  
Weight of dish and ash = 132.05 gms  
*a. 38.3%  
b. 66.4%  
c. 42.6%  
d. 58.0%  
85. A device called an Imhoff cone is commonly used to measure settleable solids in:  
a. %  
*b. mL/L  
c. mg/L  
d. ppm  
e. SVI units  
86. An aerobic treatment process is one that requires the presence of:  
a. Ozone  
b. organic oxygen  
c. no oxygen  
d. combined oxygen  
*e. dissolved oxygen  
87. A pH probe:  
a. Can be used to measure ORP in chlorine disinfection.  
b. Is often used to measure hydrogen production in wet wells.  
c. Measures, in millivolts, the difference between oxidants like chlorine and reductants such as  
organic matter.  
66  
Chapter 4. Constituents Properties and Analysis  
*d. Measures hydrogen ion activity in wastewater.  
e. Sends a 4-20 mA signal directly to a chlorine controller.  
88. Sludge solids in wastewater has an average specific gravity of 1.2; this means they are  
a. 12% heavier than water  
*b. 20% heavier than water  
c. 2% lighter than water  
d. 20% lighter than water  
89. How should the pH electrode be stored when not in use:  
a. In a strong acid solution  
b. In a strong caustic solution  
c. In a safe place in a drawer  
*d. In distilled water  
e. In a detergent  
90. In the normal Winkler test:  
a. A snow white precipitate forms in direct proportion to the nitrate concentration  
b. A brownish flocculant precipitate is evidence that D.O. is absent  
c. An endpoint is reached when a dark blue color changes to black  
d. The muffle furnace must be in excess of 500 deg. C before incubation  
*e. A snow white precipitate forms if DO is absent  
91. Organisms in wastewater that are not harmful to humans but are indicators of diseases are:  
a. Pathogens  
b. Viruses  
*c. Coliform  
d. Bacteria  
92. The typical range of suspended solids in domestic influent wastewater is:  
*a. 100-300 mg/L  
b. 400-600 mg/L  
c. 700-900 mg/L  
d. 1000-12000 mg/L  
93. Which of the flowing statement(s) is/are true with regards to BOD  
i) BOD test results are suitable for quickly establishing process efficiencies  
ii) BOD value is always greater than the COD value of the same wastewater sample  
iii) BOD is expressed in mg/ L or in ppm  
iv) BOD is the measure of organic strength  
v) BOD stands for biological oxygen demand  
a. i) & ii)  
b. i), ii) & iv)  
c. i), iv) & v)  
*d. iii) & iv)  
e. iii), iv) & v)  
94. An amperometric titrater is used to measure  
a. Alkalinity  
*b. Chlorine residual  
c. Conductivity  
d. COD.  
4.4 Laboratory Analysis  
67  
95. Which of these pH readings indicates an acidic wastewater?  
*a. 3  
b. 7  
c. 9  
d. 12  
96. Which of the following conditions will probably cause the greatest change in pH?  
a. Buffering sample  
*b. Exposing sample to atmosphere  
c. Fixing sample  
d. Refrigerating sample  
97. Nderlinehspace1cm matter in wastewater, is normally composed of grit, sand and silt.  
a. Colloidal  
*b. Inorganic  
c. Organic  
d. Volatile  
98. Which of the following characteristics would be least helpful to an operator assessing the organic  
loading on his plant?  
a. solids concentration  
b. BOD  
c. COD  
*d. pH  
e. nitrogen content  
99. Pathogens  
*a. Are bacteria or virus that cause disease.  
b. Are bacteria which do not occur in water.  
c. Can obtain their food supply without help.  
d. Are not harmful to man.  
100. Regarding the total coliform test which one of the following statements is TRUE?  
a. If less than 2 coliforms per 100 mL are found in a secondary effluent, we can be assured that we  
have destroyed all viruses and coliforms.  
b. Total coliforms are monitored in wastewater because they are generally considered to be disease  
causing organisms.  
c. Sodium thiosulfate should be added drop-wise after collection of a coliform sample to destroy  
residual chlorine.  
*d. The multiple tube fermentation method for measuring total coliforms is only a statistical esti-  
mate of the coliform organism concentration.  
101. Results from the Multiple-Tube Fermentation Technique for members of the Total Coliform Group  
are expressed as  
a. DPD.  
b. MF.  
c. MGD.  
*d. MPN.  
102. The advantage in the use of coliform organisms as an indicator lies in the following fact:  
a. They are found everywhere and they grow in common bacterial media  
b. They are found everywhere and are readily killed by chlorine  
68  
Chapter 4. Constituents Properties and Analysis  
*c. They are predominant bacteria associated with intestinal discharges and grow on  
nutrient agar forming characteristic colonies  
103. The purpose of adding sodium thiosulfate to a microbiological sample bottle is to  
a. Extend the allowable holding time from 6 to 30 hours.  
b. React with nitrates that interfere with the MPN test.  
*c. Remove any chlorine residual present.  
d. To ensure sterilization of sample bottle.  
104. The solids in a raw wastewater sample may be classified in several different categories. However,  
one may say that all the solids in a wastewater sample may be expressed as the sum of:  
a. The total of settleable solids + total dissolved solids.  
*b. The total dissolved solids + total suspended solids.  
c. The total of settleable solids + colloidal solids + total suspended solids  
105. The term MPN is used in reference to:  
a. The mass of phosphorus and nitrogen per unit of carbon  
*b. The number of coliforms per unit volume of sample that is most likely to have caused the  
observed results in a multiple tube test  
c. The number of fecal stetococci per unit volume of sample that is most likely to have caused the  
observed results in the multiple tube test  
d. The result of membrane filter test  
e. The standard plate count result  
106. The volatile portion of suspended solids contained in normal domestic wastewater could be ex-  
pected to be in the range of  
a. Less than 10 b. 25-50*c. 70-80d. 90-100  
107. What disease is not considered to be normally conveyed or transmitted by untreated wastewater?  
a. Amoebic dysentery.  
b. Hepatitis.  
*c. Malaria.  
d. Chlorea.  
108. What test is not run on the influent?  
a. BOD.  
*b. Fecal coliform.  
c. Suspended solids.  
d. pH.  
109. Which one of the following statements is TRUE regarding the standard BOD5 test?  
a. Some NPDES permits specify that only CBOD is to be reported on a wastewater plant’s final  
effluent. The letters CBOD refer to complete BOD (i.e the TOTAL BOD).  
b. Phenylarsineoxide (PAD) is typically added to destroy nitrifying bacteria when running this test.  
c. Dechlorinated secondary effluents need not be seeded when set up for the BODs test.  
d. Nitrate ions interfere with this test  
*e. The BOD measured includes the nitrogenous BOD (nBOD).results due to the oxygen demand  
exerted by certain bacteria as they oxidize ammonia to nitrate ions.  
110. A composite sample will provide a(an)  
a. Even color.  
b. High pH.  
c. Low solids sample.  
4.4 Laboratory Analysis  
69  
*d. Representative sample.  
111. Flow proportionate composite samples are collected because:  
a. The waste characteristics are continually changing  
b. The flow is continually changing  
*c. The flow and waste characteristics are continually changing  
d. This requires less time than grab samples  
e. All of the above  
112. Over a four-year period, the totalizing hour meter on an instrument air compressor had the follow-  
ing readings at the end of each year: 1st year - 9,763; 2nd year - 13,258; 3rd year - 20,071; and 4th  
year - 23,714 How many hours does the meter show the compressor ran during  
the third year?  
a. 349.5 hours  
b. 364.3 hours  
*c. 681.3 hours  
d. 830.2 hours  
113. Grab sample is always collected for which of the following test  
a. BOD  
b. TSS  
*c. Coliform  
d. COD  
e. None of the above  
114. Samples collected over several hours during the day and combined are known as:  
*a. Composite samples.  
b. Grab samples.  
c. Deep samples.  
d. Periodic samples.  
115. Grab samples are considered to be representative of the  
a. Average daily condition at the sample location  
b. Average daily condition in the system  
c. System conditions for the two hours before and after the sample was taken  
*d. System condition at the time of the sample  
116. Characteristics that should be measured immediately after the sample is collected are:  
a. Velocity and dissolved solids  
*b. Temperature, pH and DO  
c. TSS and BOD  
d. Hardness and alkalinity  
117. A stilling well on the effluent side of a wastewater facility is:  
a. a chlorine injection concentrator  
b. an automatic sampler for coliform counts  
*c. a structure containing the float for a flow measuring device  
d. dry well side of a pump station  
e. a none of the above  
118. The advantages of automatic sampling equipment are:  
a. elimination of human error inherent in manual sampl ing  
b. reduction of personnel requirements and cost  
70  
Chapter 4. Constituents Properties and Analysis  
c. allows for more frequent sampling  
d. collection of more representative samples  
*e. all of the above  
119. In collecting a sample for a chlorine residual determination of the final effluent, the most suitable  
sampling point is:  
a. at the site of chlorine injection  
b. at the entrance to the chlorine contact chamber  
c. at the midpoint of the chlorine contact chamber  
d. at the exit side of the chlorinator  
*e. at the point of effluent discharge  
120. The recommended minimum portion which should be collected for testing in sampling wastewater  
is: (Assume a grab sample)  
a. 10 ml  
b. 50 ml  
c. 10 0 ml  
d. 500 ml  
*e. 1,000 ml  
121. A composite sample will give a(n)  
a. Even color  
b. High pH  
c. Instantaneous sample  
*d. Representative sample  
122. Chlorine residual may be determined using the reagent  
*a. Diethyl-p-phenylenediamine (DPD)  
b. Ethylendiamine tetraacetic acid (EDTA)  
c. Polychlorinated biphenyls (PCB)  
d. Sodium thiosulfate (Na2S203)  
123. BOD5 is the most common method to quantify the organic content in wastewater. Another method  
used is:  
a. Chemical oxygen demand  
b. Volatile solids analysis  
c. Total organic carbon  
*d. Any of the above  
124. The laboratory test results on domestic raw sewage were: COD = 320 mg/1 BOD = 475 mg/1. The  
best interpretation would be:  
a. the raw wastewater had a high grease content  
b. the raw wastewater was septic  
*c. the laboratory results, as reported, were in error  
d. the sample was held too long before analysis  
e. the glass fiber filters used to run the test were contaminated  
125. TKN is the measure of:  
a. Ammonia/Ammonium+Inorganic Nitrogen  
*b. Ammonia/Ammonium+Organic Nitrogen  
c. Nitrate+Nitrite+Organic Nitrogen  
d. Organic Nitrogen+Inorganic Nitrogen  
4.4 Laboratory Analysis  
71  
126. Volatile solids concentration of sludge with 6  
*a. 42,000mg/l  
b. 70,000mg/l  
c. 42mg/l  
d. 700mg/l  
127. Wastewater solids can be categorized as:  
a. Suspended+Fixed  
b. Dissolved+Volatile  
*c. Volatile+Fixed  
d. Volatile+Settleable  
128. Non-settleable solids are composed of:  
a. Volatile +Dissolved  
b. Dissolved+Settleable  
c. Floatable + Suspended  
*d. Collodial + Floatable  
129. Sludge with a specific gravity of 1.1 will weigh:  
a. 1.1lbs/gal  
b. 8.34lbs/gal  
*c. 9.17lbs/gal  
d. 7.58lbs/gal  
130. A composite sample will provide a(an)  
a. Even color.  
b. High pH.  
c. Low solids sample.  
*d. Representative sample.  
131. Bacteria which cause diseases in man are generally called:  
a. Mesophilic  
b. Facultative  
*c. Pathogenic  
d. Coliforms  
132. How should the pH electrode be stored when not in use:  
a. In a strong acid solution  
b. In a strong caustic solution  
c. In a safe place in a drawer  
*d. In distilled water  
e. In a detergent  
133. In the normal Winkler test:  
a. A snow white precipitate forms in direct proportion to the nitrate concentration  
b. A brownish flocculant precipitate is evidence that D.O. is absent  
c. An endpoint is reached when a dark blue color changes to black  
d. The muffle furnace must be in excess of 500 deg. C before incubation  
*e. A snow white precipitate forms if DO is absent  
134. The typical range of suspended solids in domestic influent wastewater is:  
*a. 100-300 mg/L  
b. 400-600 mg/L  
72  
Chapter 4. Constituents Properties and Analysis  
c. 700-900 mg/L  
d. 1000-12000 mg/L  
135. Which of the flowing statement(s) is/are true with regards to BOD  
1) BOD test results are suitable for quickly establishing process efficiencies  
2) BOD value is always greater than the COD value of the same wastewater sample  
3) BOD is expressed in mg/ L or in ppm  
4) BOD is the measure of organic strength  
5) BOD stands for biological oxygen demand  
a. 1) and 2)  
b. 1), 2) & 4)  
c. 1), 4) & 5)  
*d. 3) & 4)  
e. 3), 4) and 5)  
136. Nderlinehspace1cm matter in wastewater, is normally composed of grit, sand and silt.  
a. Colloidal  
*b. Inorganic  
c. Organic  
d. Volatile  
137. Pathogens  
*a. Are bacteria or virus that cause disease.  
b. Are bacteria which do not occur in water.  
c. Can obtain their food supply without help.  
d. Are not harmful to man.  
138. Regarding the total coliform test which one of the following statements is TRUE?  
a. If less than 2 coliforms per 100 mL are found in a secondary effluent, we can be assured that we  
have destroyed all viruses and coliforms.  
b. Total coliforms are monitored in wastewater because they are generally considered to be disease  
causing organisms.  
c. Sodium thiosulfate should be added drop-wise after collection of a coliform sample to destroy  
residual chlorine.  
*d. The multiple tube fermentation method for measuring total coliforms is only a statistical esti-  
mate of the coliform organism concentration.  
139. Results from the Multiple-Tube Fermentation Technique for members of the Total Coliform Group  
are expressed as  
a. DPD.  
b. MF.  
c. MGD.  
*d. MPN.  
140. The advantage in the use of coliform organisms as an indicator lies in the following fact:  
a. They are found everywhere and they grow in common bacterial media  
b. They are found everywhere and are readily killed by chlorine  
*c. They are predominant bacteria associated with intestinal discharges and grow on  
nutrient agar forming characteristic colonies  
141. The purpose of adding sodium thiosulfate to a microbiological sample bottle is to  
a. Extend the allowable holding time from 6 to 30 hours.  
4.4 Laboratory Analysis  
73  
b. React with nitrates that interfere with the MPN test.  
*c. Remove any chlorine residual present.  
d. To ensure sterilization of sample bottle.  
142. The term MPN is used in reference to:  
a. The mass of phosphorus and nitrogen per unit of carbon  
*b. The number of coliforms per unit volume of sample that is most likely to have caused the  
observed results in a multiple tube test  
c. The number of fecal stetococci per unit volume of sample that is most likely to have caused the  
observed results in the multiple tube test  
d. The result of membrane filter test  
e. The standard plate count result  
143. Describe total coliform & fecal coliform bacteria. What is the difference between the two? De-  
scribe the test procedures for both.  
Response:  
a Describe total coliform & fecal coliform bacteria.  
Coliform bacteria are a broad group of bacteria found in soil, water and other environ-  
ments.  
• Fecal coliform are coliforms which originate in the intestines of warm-blooded animals  
b What is the difference between the two?  
While total coliforms are found widely in different environments, fecal coliforms are typi-  
cally found in the intestines of warm blooded animals and are indicators of fecal contamina-  
tion.  
c Describe the test procedures for both.  
The Multiple Tube Fermentation method to estimate the quantity of both these microorgan-  
isms includes the following steps:  
Conducting a presumptive test by inoculating the sample in a set of 15 tubes containing  
Lauryl Tryptose Broth each containing an inverted Durham tube followed by incubation  
and observation of a positive result for each tube indicated by turbidity and presence of  
gas bubble.  
Conducting a confirmed test by innoculating each of the positive samples from the  
presumptive test into a tube containing BGB broth - for Total Coliform and EC Broth -  
for Fecal Coliform  
Conducting a completed test by streaking and incubating an agar plate with positives  
from the confirmed test followed by innoculating and incubating an agar slant and  
nutrient broth with colonies from the agar plate.  
144. What is ORP Response:  
Oxidation is a chemical reaction in which electrons are gained by oxidizing agent and lost by the  
substance being oxidized. Electrons cause bonds to be broken in the organic matter thus destroying  
it.  
Oxidation potential is the direct measure in millivolts of demand for chlorine; more chlorine in-  
creases oxidants potential more organic matter lowers oxidation potential.  
Using ORP probe reduces amount of chlorine used for disinfection. Chlorine by-products are toxic  
cause problems in effluent toxicity test.  
74  
Chapter 4. Constituents Properties and Analysis  
ORP (oxidation reduction potential) probes have been found to be effective in precisely controlling  
chlorine disinfection as well as de-chlorination. An ORP probe measures directly the oxidation  
potential and compares (his measured potential to a set point (typically 540 mV). A 4-20 milliamp  
controller can be used to adjust the chlorine feed to use precisely the amount of chlorine needed for  
disinfection.  
145. When chlorine is added to a wastewater effluent it is said to act as an oxidizing agent. Define  
oxidation. Explain how oxidation might be effective in disinfection of a wastewater effluent.  
Response:  
Oxidation is a chemical reaction in which electrons are gained by the oxidizing agent and lost by  
the substance being oxidized.  
These lost electrons cause bonds to be broken in the organic matter (bacteria) and thus destroy it.  
146. Define oxidation potential. Explain how (his potential changes in relationship to the addition of  
chlorine or the increase in organic material or bacteria in the wastewater.  
Response:  
Oxidation potential is a direct measure, in millivolts, of the potential for electron transfer to the  
oxidizing agent (chlorine).  
More chlorine increases oxidation potential.  
More organic material or bacteria lowers the oxidation potential.  
147. One WWTP found that it was better able to meet its effluent toxicity limit when using an ORP  
probe. Identify and briefly explain one possible reason for this.  
Response:  
Using an ORP probe reduces significantly the amount of chlorine being used for disinfection.  
Chlorine by-products are often toxic and thus cause problems in the effluent toxicity test (bioassay  
test).  
Reducing the chlorine used may make it easier to meet the toxicity limit.  
148. An ORP probe must be place several minutes downstream of the point where chlorine is added in a  
chlorine contact tank. Why? Identify the data and conversion factor(s) necessary to calculate where  
an ORP probe is to be placed. Assume that at peak dry flow (he probe is to be placed 5 minutes  
downstream of the point where chlorine is added. Show the steps necessary to do this calculation.  
Response:  
Reactions with ammonia-nitrogen are not instantaneous. Thus, sometime between the point of  
addition of chlorine and the site of the ORP probe to allow these reactions to be complete. Data  
needed to calculate position of the probe: Peak Flow (MGD), the conversion factor = MGD =  
1.547 cu. ft. / sec., the width of the chlorine contact chamber, the depth of the water in the chlorine  
contact chamber (so that the cross-sectional area of the flow can be calculated), 60 sec = 1 minute.  
1. Find the velocity of flow (ft/sec): Q at peak x 1.547 cu. ft./sec divided by cross-sectional area in  
ft2. 2. Convert velocity (ft/sec) to ft/min = ft/sec x 60 sec/min = ft/min 3. Multiply ft/min x 5 min  
to find ft.  
5. Collections  
The collection system resembles a tree that branches out from the treatment plant to collect the wastewater  
from individual sources.  
5.1 Wastewater Collection Piping  
• A lateral is the piping that connects the public sewer to the building.  
• Laterals flow into larger lines called mains.  
• Mains carry the flow into the largest lines in the system, called trunk lines.  
• A trunk line is the pipe that brings water into the treatment plant.  
5.2 Sanitary Sewer Systems  
Sanitary sewer systems collect and convey wastewater from residential, commercial and industrial sources  
to a centralized wastewater treatment facility for treatment.  
5.2.1 Storm-water systems  
Storm-water systems are designed solely for the conveyance of storm-waters waters directly to streams,  
rivers, lakes, or the ocean.  
5.2.2 Combined sewer systems  
Combined sewer systems collect and convey sanitary sewage and urban runoff in a common piping  
system.  
Combined sewers could potentially cause serious water pollution problems during combined sewer  
overflow (CSO) events when wet weather flows exceed the sewage treatment plant capacity.  
 
 
 
 
 
76  
Chapter 5. Collections  
Combined System  
Separated System  
5.3 Collections Systems Basics  
The primary type of a collection system is a gravity system. A gravity system is so named because  
the wastewater flows down gradient in the sewer, driven by forces of gravity.  
The collection system includes the gravity sewers, force mains, manholes, pumping equipment, and  
other facilities that collect and convey the water to a wastewater treatment plant.  
Sewers are generally laid at a minimum slope to ensure open channel flow through the pipe at a  
minimum velocity of 2.0 feet per second. The minimum velocity is required to ensure that solids do  
not settle out in the sewer.  
When the sewer lines reach a certain depth, the flow must be lifted back through a lift or pump  
station.  
Lift stations are built whenever wastewater must be pumped to a higher altitude, whether it’s to lift  
water up so that it can gravity flow or to pump it over a rise or hill.  
The discharge from the pump station may be to another gravity sewer at that location or through a  
pressurized force main.  
Key elements of lift stations include a wastewater receiving well (wet-well), pumps and piping with  
associated valves.  
The size of the wet well affects the operating of the station. If a wet well is too small, excessive  
starting and stopping of the pump motors will occur, resulting in premature failure. If the wet well  
is too large, solids will tend to settle on the bottom, blocking the pump suction line and leading to  
the generation of hydrogen sulfide and methane.  
The dry well is the portion of the dry well/wet well pumping station that houses the necessary  
equipment required to pump the wastewater. The dry well is so named because it is isolated from  
the incoming wastewater.  
• Centrifugal pumps are the most common type of pump found in wastewater pumping stations.  
• In the USA, wastewater generated in a typical home is about 70 gal/day/person  
5.4 Collections Related Operational Issues  
Infiltration and inflow (I/I) is the unwanted flow into the wastewater collections systems.  
 
 
5.4 Collections Related Operational Issues  
5.4.1 Infiltration  
77  
Groundwater entering sanitary sewers through defective pipe joints and broken pipes is called  
infiltration. Remember: ground filters  
5.4.2 Inflow  
During rainstorms or snow thaws, large volumes of water may flow into the wastewater collections  
systems through leaky manhole covers or combined storm-water /wastewater connections. In  
addition, private residences may have roof, cellar, yard, area, or foundation drains inappropriately  
connected to sanitary sewers. These flows are termed as inflows. Remember: rain flow  
Implications of I/I:  
• I/I decrease the efficiency and capacity of wastewater collection systems and treatment systems  
• I/I can advance the need for capital costs to manage and treat flows  
I/I contribute to the hydraulic overloading of treatment processes, which can affect public health  
and the compliance with NPDES permit requirements  
I/I could cause Sanitary and combined sewer overflows(SSOs and CSOs) when wastewater flow  
volumes exceed the design capacity of the treatment plant  
• I/I increase collection system and treatment facility operating costs  
5.4.3 Odors  
• Hydrogen sulfide (H2S), its associated compounds and methane are generated due to microbial  
activity in wastewater and the conveyance systems. The rotten egg like smell of hydrogen sulfide  
causes public nuisance odors, poses a health hazard for collections systems workers and causes  
corrosion of the system through its conversion to sulfuric acid. The hydrogen sulfide generation is  
typically controlled by reducing the potential for septicity in wastewater through proper design of  
the system - adequate velocities and adequate air space and through chemical treatment.  
5.4.4 Fats, Oils and Grease (FOG)  
FOG from food home, food establishments and industries affect the operation of the collection  
system  
• FOG has a tendency to accumulate in sewer pipes decreasing its wastewater carrying capacity.  
• Excessive FOG accumulation may cause sewer overflows  
 
 
 
 
Chapter Assessment  
1. Hydrogen sulfide gas in a moist atmosphere can result in corrosion of concrete structures  
*a. True  
b. False  
2. As a rule of thumb, the velocity in a gravity sewer should be at least 2 cubic feet per second to  
ensure that the solids do not settle out in the sewer lines  
a. True  
*b. False  
3. A combined sewer implies that it carries domestic and industrial wastes only  
a. True  
*b. False  
4. Hydrogen sulfide gas in a moist atmosphere can result in corrosion of concrete structures  
*a. True  
b. False  
5. Inflow is storm water entering into the sewer system  
*a. True  
b. False  
6. A combined wastewater collection system handles only domestic waste and industrial waste.  
a. True  
*b. False  
7. Fats, oils and grease accumulation in sewers can cause sewer overflows  
*a. True  
b. False  
8. Storms could potentially cause sewer overflows in a Combined Sewer system  
*a. True  
b. False  
9. Groundwater entering sewer collection pipes through cracks and defective pipe joints is termed as  
80  
Chapter 5. Collections  
inflow  
a. True  
*b. False  
10. A Combined sewer system is the one that brings the combined domestic and industrial wastewater  
flow to the treatment plant  
a. True  
*b. False  
11. Infiltration is when groundwater enters the sewage collections systems  
*a. True  
b. False  
12. Infiltration is when groundwater enters the sewage collections systems  
*a. True  
b. False  
13. As a rule of thumb, the velocity in a gravity sewer should be at least 2 cubic feet per second to  
ensure that the solids do not settle out in the sewer lines  
a. True  
*b. False  
14. A combined sewer implies that it carries domestic and industrial wastes only  
a. True  
*b. False  
15. Hydrogen sulfide gas in a moist atmosphere can result in corrosion of concrete structures  
*a. True  
b. False  
16. Inflow is storm water entering into the sewer system  
*a. True  
b. False  
17. A lateral is the largest sewer line which brings the wastewater to the treatment plant  
a. True  
*b. False  
18. Hydrogen sulfide gas in a moist atmosphere can result in corrosion of concrete structures  
*a. True  
b. False  
19. As a rule of thumb, the velocity in a gravity sewer should be at least 2 cubic feet per second to  
ensure that the solids do not settle out in the sewer lines  
a. True  
*b. False  
20. A combined sewer implies that it carries domestic and industrial wastes only  
a. True  
*b. False  
21. Hydrogen sulfide gas in a moist atmosphere can result in corrosion of concrete structures  
*a. True  
b. False  
22. Inflow is storm water entering into the sewer system  
*a. True  
b. False  
5.4 Collections Related Operational Issues  
81  
23. A combined wastewater collection system handles only domestic waste and industrial waste.  
a. True  
*b. False  
24. Fats, oils and grease accumulation in sewers can cause sewer overflows  
*a. True  
b. False  
25. Storms could potentially cause sewer overflows in a Combined Sewer system  
*a. True  
b. False  
26. Groundwater entering sewer collection pipes through cracks and defective pipe joints is termed as  
inflow  
a. True  
*b. False  
27. A Combined sewer system is the one that brings the combined domestic and industrial wastewater  
flow to the treatment plant  
a. True  
*b. False  
28. Infiltration is when groundwater enters the sewage collections systems  
*a. True  
b. False  
29. Define Infiltration and Inflow. Discuss the impact of inflow and infiltration on the wastewater  
treatment plant:  
30. In wastewater collections what is a gravity system:  
31. A sewer system designed to transport only wastewater from homes, industries, institutions and  
businesses is called:  
a. Combined sewer  
*b. Sanitary sewer  
c. Service sewer  
d. Building sewer  
32. Velocity of sewers is usually expressed as:  
*a. Feet per second  
b. Gallon per minute  
c. MGD  
d. mg/l  
e. sq. ft  
33. Which of the following statements is not true regarding a wastewater collection system:  
*a. A sewer is designed to allow the waste to flow at a rate of approximately 1 ft/sec.  
b. Grease can be serious problem in a collection system.  
c. Inflow and infiltration are frequently problems in older collection systems.  
d. High concentrations of hydrogen sulfide in a sewer can lead to corrosion of concrete.  
e. Scouring can be a problem if wastewater is flowing too fast.  
34. Which of the following statements is not true regarding a wastewater collection system:  
*a. A sewer is designed to allow the waste to flow at a rate of approximately 1 ft/sec.  
b. Grease can be serious problem in a collection system.  
c. Inflow and infiltration are frequently problems in older collection systems.  
82  
Chapter 5. Collections  
d. High concentrations of hydrogen sulfide in a sewer can lead to corrosion of concrete.  
e. Scouring can be a problem if wastewater is flowing too fast.  
35. A sewer system designed to transport only wastewater from homes, industries, institutions and  
businesses is called:  
a. Combined sewer  
*b. Sanitary sewer  
c. Service sewer  
d. Building sewer  
36. Velocity of sewers is usually expressed as:  
*a. Feet per second  
b. Gallon per minute  
c. MGD  
d. mg/l  
e. sq. ft  
37. Velocity of flow in sewers is usually expressed in terms of  
*a. Feet per second  
b. Gallon per minute  
c. MGD  
d. Milligrams per liter  
e. Square feet  
38. Velocity of flow in sewers is usually expressed in terms of  
*a. Feet per second  
b. Gallon per minute  
c. MGD  
d. Milligrams per liter  
e. Square feet  
39. Infiltration is caused by: *a. Cracked pipes b. Improper CCTV operation c. Poor ventilation d. All  
of the above  
40. Define infiltration & inflow, compatible & non-compatible. List 4 major concerns. List 2 correc-  
tions.  
6. Preliminary Treatment  
The objective of preliminary treatment is to remove coarse solids and other large materials often  
found in raw wastewater  
Removal of these materials is necessary to enhance the operation and maintenance of subsequent  
treatment units  
• Preliminary treatment operations typically include a combination of the following processes:  
Screening  
Grinding or shredding  
Flow measurement  
Grit removal  
Pre-aeration  
Flow equalization  
6.1 Process Elements of Preliminary Treatment  
6.1.1 Screening  
• Screening is typically the first unit in a preliminary treatment  
Screening allows for the capture of coarse solids as pieces of cloths garbage so as to protect pumps  
and other units from clogging.  
Screens may consist of vertical or inclined bars (bar racks or bar screens), wire mesh or perforated  
plates having either circular or rectangular openings.  
Screens remove the large, entrained, suspended or floating solids such as pieces of wood, cloth,  
paper, plastics, garbage, etc.  
• Debris collected on the screen can be cleaned manually or automatically using chain driven rakes  
• The retained material at screens - screenings, is collected and hauled to landfill for disposal  
The quantity of screenings removed varies by location and is a function of the clear opening of the  
screen.  
 
 
 
84  
Chapter 6. Preliminary Treatment  
Barscreen - No rakes  
Barmuinitors combine the function of a screen and a grinder. The ground material is returned to the  
wastewater flow for removal during primary treatment.  
6.1.2 Grinding and Shredding  
Comminutor(Grinder) consist of fixed, rotating or oscillating teeth or blades, acting together to reduce the  
solids to a size which will pass through fixed or rotating screens grind rags into small chunks  
The comminutors are installed in wastewater channel and they grind the larger solids without actually re-  
moving them from the wastewater. These devices may be installed before the screens or as a combination  
of screen and cutters (barmunitors).  
Comminutor  
6.1.3 Flow Measurement  
Wastewater flow to a treatment plant is not constant but varies in a diurnal (daily) pattern reflecting  
domestic water use activity.  
 
 
6.1 Process Elements of Preliminary Treatment  
85  
Schematic of comminutor placement in a channel  
Continuous flow measurement is necessary in order to monitor diurnal variations in flow which  
may affect treatment plant efficiency.  
Devices used for flow measurement as part of the preliminary treatment can be placed in a channel  
or in a pipe.  
Devices for Flow Measurement in Channels  
• Weirs:  
Typically sharp crested weirs which are essentially metal plates installed perpendicular the  
flow. The plate may a straight edge, a V-notch, or a trapezoidal opening.  
The weir plate in the channel causes an increase in the depth of the water behind the weir.  
which is proportional to the flow rate.  
The flow rate can be determined by calculation or by reading the corresponding flow value to  
the depth of the water, from a chart specific for that weir.  
(a) Weir in a channel  
• Parshall Flume:  
(b) Weir types  
Parshall flume uses a narrow section in the channel as the restriction rather than the vertical  
plate of a weir.  
86  
Chapter 6. Preliminary Treatment  
Like the weir, the restriction due to the narrow section of the flume, causes an increase in the  
depth of the water behind the weir (head) which is proportional to the flow rate.  
The flow rate can be determined by calculation or by reading the corresponding flow value to  
the depth of the water, from a chart specific for that Parshall Flume.  
(a) Parshall flume  
(b) Parshall flume with level sensor  
Devices for Flow Measurement in Pipes  
Venturi Tube:  
Measures the difference in pressure in the inlet and center section (throat)  
This pressure difference can then be mathematically converted to a flow rate.  
Works only when a pipe if flowing full  
• Magnetic Flow Meter (Magmeter):  
Magmeter is a pipe spool which has an electromagnetic coil surrounding it. As the wastew-  
ater - a conducting material, flows through it, an electrical current is created proportional to  
the velocity of the conducting fluid (wastewater).  
Flow is automatically calculated by multiplying the velocity by the cross- sectional area of  
the pipe.  
Similar to the venturi meter, magmeter will read accurately only if the magmeter section  
of the pipe is flowing full and the wastewater is flowing through it at a certain minimum  
velocity.  
(a) Magmeter  
(b) Piping with magmeters  
6.1 Process Elements of Preliminary Treatment  
87  
6.1.4 Grit Removal  
Grit includes sand, gravel, cinder, eggshells, bone chips, seeds, coffee grounds, and large organic  
particles, such as food waste.  
• Purpose of Grit removal:  
to protect mechanical equipment from abrasion and abnormal wear  
to reduce clogging caused by deposition of grit particles in pipes and channels, and  
to prevent loading the treatment plant with inert matter that might interfere with the operation  
of treatment units such as anaerobic digester and aeration tanks.  
• Removal of organic material along with the grit is undesirable for two reasons:  
1. It causes odor issues, and  
2. Organic matter is a potential source of energy (digester gas)  
• Grit Disposal: Grit removed is typically landfilled.  
• Grit Volume: The volume of grit collected measured in ft3/MG.  
• The rate of grit collection can range from 0.5 ft3/MG to 30 ft3/MG.  
Wastewater plants having a combined collection system must deal with much larger volumes of  
grit.  
Grit Removal Systems  
• HORIZONTAL GRIT CHAMBERS:  
These are rectangular channels 30 to 60 feet long and the water detention time is between 45  
to 90 seconds  
Water passing through these channels is maintained at a relatively constant velocity of about  
1 feet per second (fps) which allows for the grit to settle while keeping the lighter organic  
material to stay in suspension and continue on into the primary clarifiers.  
• AERATED GRIT CHAMBERS:  
The 1 fps velocity is maintained by using aerators to create a rolling flow in the tank.  
Aeration is achieved using diffusers located on the bottom of one side of the grit chamber.  
Aerated grit chambers help create aerobic conditions in septic sewage. Aerobic conditions  
help improve the settleability of the sludge and increase both BOD and suspended solids  
removal in the primary clarifiers.  
Much larger and deeper than non-aerated units.  
The detention times are increased to 3 to 5 minutes.  
(a) Horizontal grit chamber  
(b) Aerated grit chamber  
• CYCLONIC/VORTEX GRIT CHAMBER:  
The wastewater flows into a cylinder that tapers to a cone at one end.  
 
88  
Chapter 6. Preliminary Treatment  
The flow whirls around the inside of the cylinder like a cyclone which causes the heavy grit  
to be slinged to the outside and it ultimately settles to the bottom from where it is withdrawn.  
(a) Vortex grit chamber design  
(b) Vortex grit chamber installed  
Grit Removal  
In the cyclonic/vortex grit chamber, the grit is scoured with water and is removed using  
pumps  
For the horizontal and aerated grit systems:  
Mechanical augers at the bottom of the grit chamber move the grit to one end of the tank  
where grit slurry pumps can pump it out of the tank to a grit separator.  
In some cases steep bottom slope is provided which will collect the grit at Central Point  
of Removal.  
*
*
Grit Removal is achieved by air pumps for small aerated grit chambers.  
Grit can also be removed by tubular conveyors, buckets type collectors, elevators screws  
conveyors, grit pumps and clam shell buckets  
*
*
6.1.5 Flow Control  
Flow control is critical for grit removal, specifically for the horizontal and aerated grit chambers  
as excessive or inadequate velocities would lead to poor grit removal or cause excessive organic  
material settling along with the grit, respectively  
As the wastewater flows vary diurnally, it is important that velocity of the wastewater in the grit  
chamber should be maintained nearly constant - near 1 fps.  
Constant velocity in a grit chamber is achieved by providing a proportional (Sutro) weir at the outlet  
end of grit chamber.  
The shape of the opening between the plates of a proportional weir is made in such a way that the  
discharge is directly proportional to liquid depth in grit chamber resulting in maintaining a constant  
velocity of water even a the flow changes.  
6.1.6 Pre-aeration  
Pre-aeration of the wastewater as part of the preliminary treatment may be provided as a separate  
process or increased detention time in an aerated grit chamber.  
• Pre-aeration provides the follwoing benefits:  
freshens up wastewater by dissolving oxygen thereby reducing the wastewater septicity  
reduction of septicity allows for better settling - solids and BOD removal, in the following  
primary treatment process  
 
 
6.1 Process Elements of Preliminary Treatment  
89  
Diurnal wastewater flow profile  
(a) Proportional weir design  
(b) Installed Proportional weir  
promotes grease separation which facilitates its removal during primary treatment  
6.1.7 Flow Equalization  
• Flow equalization involves storing a portion of peak flows for release during low-flow periods  
It prevents surges and allows for the operation of processes at design flows thus allowing for opti-  
mal physical, biological and chemical processes to take place.  
It results in saving capital costs as the processes may be built with a treatment capacity which is  
less than the peak flows  
 
Chapter Assessment  
1. A weir can be also be used for measuring flows  
*a. True  
b. False  
2. The process of pre-aeration in no way influences the degree of settling in a primary clarifier  
a. True  
*b. False  
3. Septic sludge has a low pH  
*a. True  
b. False  
4. A Parshall flume measures the velocity of the influent flow  
a. True  
*b. False  
5. A barminutor frequently operates automatically.  
*a. True  
b. False  
6. A grit chamber with a faster flow velocity than recommended may allow appreciable organic matter  
to collect in the grit.  
a. True  
*b. False  
7. A grit chamber with a slower flow velocity than recommended may allow appreciable organics to  
92  
Chapter 6. Preliminary Treatment  
settle out with the grit.  
*a. True  
b. False  
8. A Parshall Flume is a device used to divide the incoming flow for equal distribution to a plant  
having more than one primary clarifier.  
a. True  
*b. False  
9. A Parshall flume measures the velocity of the influent flow  
a. True  
*b. False  
10. A properly operated grit chamber will normally increase the solids loading to the primary clarifier  
*a. True  
b. False  
11. A properly operating grit chamber should yield grit that is high in fixed solids (inorganic material)  
and low in volatile solids.  
*a. True  
b. False  
12. At most treatment plants preliminary treatment is used to protect pumping equipment and to facili-  
tate subsequent treatment processes.  
*a. True  
b. False  
13. A Venturi meter measures the amount of electricity used and should be read when there is a high  
electrical demand.  
a. True  
*b. False  
14. A weir can be used as a flow measuring device  
*a. True  
b. False  
15. Hydrogen sulfide gas in a moist atmosphere can result in corrosion of concrete structures  
*a. True  
b. False  
16. Inefficient grit removal would tend to cause a decrease in percent volatile solids in raw primary  
sludge  
*a. True  
b. False  
17. pH value less than 7 indicates an alkaline or basic condition  
a. True  
6.1 Process Elements of Preliminary Treatment  
93  
*b. False  
18. Poor grit removal would affect all of the following: pumps and other mechanical equipment, anaer-  
obic digestion, percent volatile and percent total solids in the raw sludge.  
*a. True  
b. False  
19. Pre-aeration improves settleability in primary clarifier  
*a. True  
b. False  
20. Pre-aeration of raw wastewater may cause better solids separation and removals  
in the primary clarifier.  
*a. True  
b. False  
21. Presence of hydrogen sulfide cannot always be detected by its characteristic odor  
*a. True  
b. False  
22. In a typical treatment facility, it is necessary to have flow meters on the influent and effluent to  
detect the loss in plant flow.  
a. True  
*b. False  
23. A Venturi meter is a reliable device for measuring flows of either treated or untreated wastewater.  
*a. True  
b. False  
24. A flow measurement device such as a constant differential meter, or the more common name,  
rotameter, is used for the measurement of liquids or gases.  
*a. True  
b. False  
25. It is often necessary to pre-chlorinate wastewater to prevent odors. When this is done, it is not  
necessary to satisfy a chlorine demand or expect a chlorine residual.  
*a. True  
b. False  
26. Inefficient grit removal would tend to cause an increase in the percent volatile solids in raw sludge.  
a. True  
*b. False  
27. Grit is composed mostly of inorganic material and organic material that is not easily biodegradable  
*a. True  
b. False  
94  
28. A weir can be also be used for measuring flows  
Chapter 6. Preliminary Treatment  
*a. True  
b. False  
29. Nderlinehspace1cm is used for controlling the flow velocity in a horizontal grit channel  
Correct Answer(s):  
a. True  
b. False  
30. Septic sludge has a low pH  
*a. True  
b. False  
31. Barminutors and comminutors are devices which cut up or shred material which is normally found  
in raw wastewater.  
*a. True  
b. False  
32. Barmunitor and comminutor are devices used for cutting up or shredding material normally present  
in raw wastewater  
*a. True  
b. False  
33. Coliform testing is typically conducted on a grab sample  
a. True  
*b. False  
34. Comminutors cut up or shred large objects normally found In raw wastewater and remove them  
from the wastewater flow.  
a. True  
*b. False  
35. Conductivity is a very useful test for assessing sea water intrusion into sewer lines  
*a. True  
b. False  
36. Fresh wastewater is characterized by a blackish color, foul and unpleasant odors with floating  
materials and suspended solids.  
a. True  
*b. False  
37. Hydrogen sulfide in addition to creating an odor nuisance can be an explosion hazard when mixed  
with air in certain concentrations.  
*a. True  
b. False  
38. Inflow is storm water entering into the sewer system  
6.1 Process Elements of Preliminary Treatment  
95  
*a. True  
b. False  
39. Percent efficiency of total solids or BOD removal is calculated using the following formula: (In-  
Out*100)/(In-(In*Out))  
a. True  
*b. False  
40. Poor grit removal would affect anaerobic digester operation  
*a. True  
b. False  
41. Pre-chlorination is frequently used to disinfect raw wastewater.  
a. True  
*b. False  
@Prechlorination is primarily for reducing septicity  
42. Solids removed from preliminary treatment are typically treated in an anaerobic digester  
a. True  
*b. False  
43. The function of a comminutor is to shred rags, paper, wood, and other large wastewater solids and  
remove the from the flow.  
a. True  
*b. False  
44. The laboratory measurement of volatile solids is a fair approximation of the organic content of the  
wastewater.  
*a. True  
b. False  
45. The size and nature of solids in the wastewater is of no significant concern to the wastewater treat-  
ment plant operator.  
a. True  
*b. False  
46. The velocity of wastewater flowing through a long channel type of grit chamber may be controlled  
by a proportional weir.  
*a. True  
b. False  
47. Total solids are made up of dissolved and suspended solids both of which contain organic and  
inorganic matter  
*a. True  
b. False  
48. Typical domestic wastewater BOD content is about 2000mg/l  
96  
Chapter 6. Preliminary Treatment  
a. True  
*b. False  
49. Wastewater with a pH of 12.0 to 14.0 would be too "acidic" for biological treatment  
a. True  
*b. False  
50.  
51.  
52.  
matter in wastewater, is normally composed of grit, sand and silt.  
a. Colloidal  
*b. Inorganic  
c. Organic  
d. Volatile  
is used for controlling the flow velocity in a horizontal grit channel  
a. Magmeter  
*b. Proportional weir  
c. Parshall flume  
d. V-notch weir  
A
is used in the wastewater treatment plant to remove debris such as large rocks, branches,  
pieces of lumber, leaves, paper, tree roots, etc. from the influent.  
a. Comminutor  
*b. Bar Screen  
c. Belt filter  
d. Grit chamber  
53. A pH probe:  
a. Can be used to measure ORP in chlorine disinfection.  
b. Is often used to measure hydrogen production in wet wells.  
c. Measures, in millivolts, the difference between oxidants like chlorine and reductants such as  
organic matter.  
*d. Measures hydrogen ion activity in wastewater.  
e. Sends a 4-20 mA signal directly to a chlorine controller.  
54. A sewer system designed to transport only wastewater from homes, industries, institutions and  
businesses is called:  
a. Combined sewer  
*b. Sanitary sewer  
c. Service sewer  
d. Building sewer  
55. Carryover of grit from the grit chamber may indicate the need to:  
*a. Increase rate of settled grit removal from the grit chamber..  
b. Decrease the operational depth of the channel.  
c. Increase the flow to the primary clarifier.  
d. Increase the air input to an aerated grit chamber.  
6.1 Process Elements of Preliminary Treatment  
97  
56. Characteristics that should be measured immediately after the sample is collected are:  
a. Velocity and dissolved solids  
*b. Temperature, pH and DO  
c. TSS and BOD  
d. Hardness and alkalinity  
57. Flow proportionate composite samples are collected because:  
a. The waste characteristics are continually changing  
b. The flow is continually changing  
*c. The flow and waste characteristics are continually changing  
d. This requires less time than grab samples  
e. All of the above  
58. Grab samples are considered to be representative of the  
a. Average daily condition at the sample location  
b. Average daily condition in the system  
c. System conditions for the two hours before and after the sample was taken  
*d. System condition at the time of the sample  
59. Grit is composed mostly of which of the following substances?  
a. Grease  
b. Colloidal solids  
c. Rubber goods  
*d. Inorganics  
e. Plastics  
60. Organisms in wastewater that are not harmful to humans but are indicators of diseases are:  
a. Pathogens  
b. Viruses  
*c. Coliform  
d. Bacteria  
61. Which of the following pollutants would be removed to the extent in an efficiently operating grit  
chamber?  
a. Egg shells  
b. Seeds  
c. Oils and grease  
*d. Sand  
e. Fixed solids  
62. Proportional weirs usually are located at:  
a. Immediately after the barscreens  
b. Primary clarifiers  
c. Aerobic digester scum boxes  
*d. Grit chambers  
e. Inside the Parshall flume  
98  
Chapter 6. Preliminary Treatment  
63. Which of the following process units is not usually considered to be a preliminary treatment unit?  
a. Grit chamber  
b. Bar screen  
c. Comminutor  
d. Bar rack  
*e. A meniscus  
64. Which of the following would be included in the pretreatment unit?  
a. preaeration  
b. grit removal  
c. screening  
d. comminutor  
*e. all of the above  
65. As grit accumulates in a recetangular channel or chamber, the velocity of the influent wastewater:  
*a. increases  
b. decreases  
c. remains constant  
d. behaves independent of inflow  
e. none of the above  
66. Grit usually contains some organic matter which decomposes and creates odors. To facilitate  
disposal without nuisance, the organic matter is removed by washing methods. Commonly used is:  
a. aeration  
b. elutriation  
c. vacuum filtration  
d. wet oxidation  
*e. none of the above  
67. The flow through rate for grit channels is usually:  
a. 20 seconds to 1.0 minute  
b. 2 feet per second  
*c. 1 foot per second  
d. 30 days, depending on temperature  
e. none of the above  
68. Given the following data, what is the most likely cause of the mechanically cleaned bar-screen  
problem?  
DATA: Normal dry weather flow  
Motor running but unit not operating  
Drive chain excessively tight  
Water differential across screen above 6 inches  
Controls on automatic.  
Pin sheared or automatic clutch tripped.  
a. Bubble tube malfunction  
b. Flow too high  
6.1 Process Elements of Preliminary Treatment  
99  
*c. Rocks lodged in screen  
d. Suction channel level too high  
69. In which of the following types of wastewater would settling occur most rapidly?  
*a. Cold wastewater  
b. Old (stale) wastewater  
c. Septic wastewater  
d. Strong wastewater  
70. Given the following data, whatis the most likely cause of  
the grit separator problem?  
DATA: ·Lower than normal flow of water and grit from apex  
High separation chamber pressure  
*a. Grit pump suction clogged.  
b. Partial obstruction near apex  
c. Stick lodged in separation chamber  
d. Vortex finder worn  
71. What is the most likely cause of an aerated grit chamber problem if the grit pump is in automatic  
mode and running, the suction valve is wide open, the pressure on discharge line is low and erratic,  
and less than normal water and grit are discharging from discharge line?  
a. Discharge check valve partially plugged.  
b. Grit classifier partially plugged.  
*c. Grit pump suction line partially plugged.  
d. Malfunctioning air supply to grit chamber causing pump to get air bound.  
72. Usually, preliminary treatment includes removal of most of the:  
a. Pathogenic bacteria.  
b. Biodegradable organics.  
c. Settleable solids.  
d. Dissolved solids.  
*e. Grit.  
73. A spray nozzle on the mechanically cleaned screen has become plugged. To ensure your safety,  
prior to entering the screen housing to repair the nozzle, you should  
a. Leave note on breaker panel of repair being made  
b. Request assistance for repair •  
*c. Turn off and lock motor control  
d. Turn off local control switch  
74. An aerobic treatment process is one that requires the presence of:  
a. Ozone  
b. organic oxygen  
c. no oxygen  
d. combined oxygen  
*e. dissolved oxygen  
100  
Chapter 6. Preliminary Treatment  
75. Which of the following should not normally be a significant part of grit  
a. Sand.  
b. Rocks  
c. Eggshells.  
*d. Fecal Matter.  
76. Samples collected over several hours during the day and combined are known as:  
*a. Composite samples.  
b. Grab samples.  
c. Deep samples.  
d. Periodic samples.  
77. Which of the following is not a characteristic of hydrogen sulfide?  
a. Foul odors  
*b. Lighter than air  
c. Toxic  
d. Corrosiveness  
e. Explosiveness  
78. The following device is used to measure the flow of wastewater:  
a. Comminutor.  
b. Comparator.  
*c. Parshall flume.  
d. Sluice gate.  
79. A device called an Imhoff cone is commonly used to measure settleable solids in:  
a. Percent  
*b. mL/L  
c. mg/L  
d. ppm  
e. SVI units  
80. The proper operation of an aerated grit removal process will:  
a. Cause material with a specific gravity of greater than 1.0 to settle.  
b. Cause sand and other non-organics to settle and keep organic material in suspension.  
c. Help to freshen stale or septic wastewater.  
*d. b and c  
81. Velocity of sewers is usually expressed as:  
*a. Feet per second  
b. Gallon per minute  
c. MGD  
d. mg/l  
e. sq. ft  
82. Which of the following statements is not true regarding a wastewater collection system:  
6.1 Process Elements of Preliminary Treatment  
101  
*a. A sewer is designed to allow the waste to flow at a rate of approximately 1 ft/sec.  
b. Grease can be serious problem in a collection system.  
c. Inflow and infiltration are frequently problems in older collection systems.  
d. High concentrations of hydrogen sulfide in a sewer can lead to corrosion of concrete.  
e. Scouring can be a problem if wastewater is flowing too fast.  
83. In order for the grit to settle in a non-aerated grit chamber, the average velocity should be kept near:  
*a. 1 ft/sec .  
b. 2 ft/sec.  
c. 5 ft/sec.  
d. 1 ft/min.  
e. 2 ft/min.  
7. Primary Treatment  
• Synonyms: primary treatment basin, primary clarifier, sedimentation basin, primaries, clarifier  
• Primary treatment is after preliminary treatment and before secondary treatment  
• Its two main objectives are:  
Remove settleable solids  
Remove floatable solids  
This is a physical process which relies on the physical properties - how heavy or light the sus-  
pended solids particles are to effect its separation  
Provides quiescent conditions for the influent wastewater for the heavier solids to settle and the  
lighter solids to float  
• Removes settleable solids and floatables  
• Settled solids are removed as sludge from the bottom of the clarifier  
• Floatable solids including oil and grease are also removed, as scum from the surface  
• The shape of the primary clarifier is either rectangular or circular  
Effective solids removal in the primary clarifiers will reduce the loading on the expensive sec-  
ondary treatment process.  
The amount of solids removed during primary treatment may be enhanced by chemical addition  
- ferric or ferrous chloride as a coagulant and anionic polymer as the flocculant. This is called  
Chemically Enhanced Primary Treatment (CEPT).  
Typical Removal Rates:  
BOD removal – 25% to 40% and about 60% with CEPT  
Suspended solids (SS) removal – 40% to 60% and about 75% with CEPT  
Settleable Solids removal - >90%  
 
104  
Chapter 7. Primary Treatment  
Cross section of a Rectangular Clarifier  
Schematic cross section of a circular clarifier  
Cross section of a circular clarifier  
7.1 Clarifier Zones  
105  
7.1 Clarifier Zones  
7.1.1 Inlet Zone  
Inlet Zone is where the water enters the end of a rectangular tank, or the center of a circular or  
square tank.  
• The Inlet Zone is designed to accomplish two objectives:  
1. Reduce the velocity (dissipate energy in the incoming water)  
2. Distribute the flow evenly  
The inlet zone is equipped with a baffle. Inlet baffle reduces the velocity of the influent flow, pre-  
vent short circuiting which could cause solids being carried over to secondary treatment.  
Circular tanks are equipped with a collar-type circular baffle that directs the water down as it  
enters the center of the tank.  
Rectangular tanks will have a plate baffle in front of the opening for the wastewater flow into  
the clarifier and another baffle just upstream - a perforated wall or a picket fence type baffle  
that spreads the water laterally across the inlet end of the tank.  
(a) Rectangular clarifier influent baffle  
(b) Circular clarifier influent baffle  
7.1.2 Settling Zone  
• This is the largest portion of the tank where solids settle.  
• The water velocity is reduced to 0.03-0.05 fps and the detention time is about 1.5 to 2 hours.  
A clarifier is said to be short circuiting if the velocity of the water is greater in some sections than  
in others. The water passing through the higher velocity region will have a reduced detention time  
and settleable solids will carry through with this water as it goes over the weir. Short circuiting is  
prevented by appropriately designing inlet baffles and weir plates (at the Outlet Zone).  
7.1.3 Sludge Zone  
• Sludge zone is the bottom of the tank where the settled sludge collects and compacts.  
Sludge blanket depth should be measured and sludge should be removed at least every shift. A  
desirable blanket depth is typically established and the sludge pumping rate and regimen is estab-  
lished to maintain that desired sludge blanket level.  
• Sludge rakes push the sludge to one end or the center of the tank so that it can be pumped out.  
The rake drive is usually equipped with a torque indicator. A shear pin in the drive shaft will break  
to prevent damage to the gearbox or drive shaft.  
 
 
 
 
106  
Chapter 7. Primary Treatment  
Failure to remove sludge often enough will result in the sludge becoming septic releasing gas  
bubbles which hinders the sludge settling and also result in causing odor problems.  
The sludge from the primary clarifiers needs to be stabilized prior to its disposal. The sludge  
(solids) from the primary clarifiers are mixed with the solids from the secondary treatment pro-  
cess and stabilized typically using a sludge digestion process.  
7.1.4 Skimming Zone  
• The skimming zone is at the surface of the tank for scum removal  
• Lighter solids and greases float to the surface of the clarifier as scum  
In Circular Clarifiers: Floating matter is skimmed by a skimmer arm that is supported by the sludge  
rake and rotates with it around the tank. The floating matter is pushed over the beach plate by the  
wipers attached to the skimmer arm and into a scum box attached to the tank wall.  
In Rectangular Clarifiers: The flights act as skimmers when the chain brings them to the surface  
and pushes the scum towards the scum troughs. The scum trough may be designed to rotate (tip)  
periodically for the scum to flow in from the water surface.  
The scum collected from the primary clarifiers is sent to the digester for treatment along with the  
sludge removed.  
Rectangular clarifier scum trough  
The flights in the rectangular clarifier are supported at the top by two parallel rails running along  
the length of the clarifier.  
There are wear plates (strips) installed at the clarifer bottom and on top of the rails to prevent the  
flights from riding directly on those surfaces. To reduce friction, the flights have a wearing shoe  
 
7.1 Clarifier Zones  
107  
attached.  
• Both the wear strip and the wearing shoe are disposable items and are replaced at fixed intervals.  
Rectangular clarifier flights  
7.1.5 Outlet Zone  
This is the part of the clarifier where the settled water leaves to go to the secondary treatment  
processes.  
A channel called the effluent launder collects the effluent flow and directs it to the primary effluent  
piping.  
Weirs are installed along the edge of the effluent launder channel to skim the water evenly off the  
surface of the tank. The most common type of effluent weir is a V-notch (or saw-tooth) weir. A  
V- notch weir is a plate that has notches, about 2-3 inches deep, cut in it every 6-8 inches. If the  
weir is clean and level, it will remove water evenly all the way around the edge of the tank. This  
minimizes the upward velocities near the effluent launder and improves removal efficiencies. If  
the weir plate is not level or part of the weir becomes clogged with slime or debris, short-circuiting  
will result because more water will pass over the low side or the clean notches of the weir. Short-  
circuiting will cause poor settling and uneven sludge blanket buildup.  
• In rectangular tanks the water leaves at the end opposite the influent.  
• In circular tanks the water leaves at the edge of the tank.  
Also, in the circular clarifiers, an effluent baffle, just upstream of the weir, is installed to prevent  
floating solids from going over the weir.  
 
108  
Chapter 7. Primary Treatment  
Circular clarifier skimmer arm, effluent baffle and v-notch weir  
Rectangular clarifier v-notch weir and launder  
7.2 Sludge Pumping  
109  
7.2 Sludge Pumping  
The sludge pumping from the clarifier must be adequate to prevent sludge from going septic. Septic  
sludges are much more difficult to thicken or de-water and cause odor issues.  
• Primary sludge normally averages 4-6% solids. Generally positive displacement pumps are used for  
primary sludge  
• The pumping cycles must be designed to provide the thickest sludge possible.  
Excessive pumping or pumping without building solids to build up leads to pumping thinner (more  
water) sludge.  
7.3 Design Parameters  
Clarifier depth – 8 to 12 feet  
Hydraulic or surface loading  
This rate is important to ensure good settleable solids removal efficiency  
It is expressed in terms of gallons per day per square foot (gpd/sq ft) of tank surface area  
Typical surface loading rate used for the design of primary clarifiers range between 300 to  
1,400 gpd/sq ft, depending on the nature of the solids and the treatment requirements. Lower  
loading rates are frequently used in small plants in cold climates. In warm regions, low rates  
may cause excessive detention which could lead to septicity.  
7.4 Advance primary treatment (APT)  
Synonyms: Advance primary treatment (APT), Chemically enhanced primary treatment (CEPT), Physical-  
chemical treatment (Phys-chem)  
7.4.1 Background  
Suspended solids present in wastewater are typically coated with bacterial slime and biological  
metabolic products which are negatively charged. A significant portion of the suspended solids in  
wastewater do not settle easily due to gravity as:  
1. the biological mass and the associated byproduct gases produced makes these particles buoy-  
ant,  
2. the negative electrostatic charges on these particles cause these particles to be in constant  
state of motion due to electrostatic repulsion  
Advance primary treatment (APT) also known as Chemically enhanced primary treatment (CEPT)  
or Physical-chemical treatment (Phys-chem), involves chemical addition to the primary influent  
flow to enhance primary treatment TSS and BOD removal efficiencies  
a normal primary treatment process typically removes 40 to 60% TSS and 25 to 40% BOD. TSS  
and BOD removal efficiencies of over 80% and 60% respectively, may be achieved by the use of  
APT.  
additional cost incurred for the chemical addition is in most cases is offsetted by the benefits which  
include:  
1. by removing more BOD in the primary treatment cost associated with secondary treatment is  
reduced  
2. primary BOD is more easy to digest than the secondary biomass thus the digester gas pro-  
duction is increased and digested solids production is lowered thus saving biosolids hauling  
 
 
 
 
110  
Chapter 7. Primary Treatment  
cost  
3. residual ferric chloride in the primary sludge provides H2S and struvite control in the solids  
treatment processes  
7.4.2 Process/mechanism  
APT is a two step chemical process:  
Coagulation  
Coagulation is the process by which the negative charge on these particles is reduced lowering the  
repulsion forces, by the use of a chemical such as ferric chloride and alum.  
The concentration of the coagulant required is dependent on the strength of the wastewater and the  
conveyance time of the wastewater.  
Typically the coagulant is added immediately after the grit chambers so the conveyance from the  
grit chambers to the primary clarifier provides adequate contact time and mixing.  
• Parameters to ensure optimal coagulation are:  
Appropriate coagulation concentration  
Adequate mixing energy, and  
Adequate contact time  
• Overdosing the coagulant will adversely effect the settleability.  
• Typical ferric chloride dosage for coagulation range from 12 to 22 mg/l.  
Flocculation  
Flocculation uses an anionic polymer - polymer which has negatively charged groups, to bridge the  
coagulated particles to a size which will settle in the primary clarifier.  
The flocculated particles are prone to shearing thus the polymer is gently folded in with the coagu-  
lated wastewater just prior to entry into the primary clarifier  
CEPT Schematic  
Untreated Primary Inluent  
Coagulation  
Flocculation  
 
Chapter Assessment  
1. An Imhoff cone is often used to measure the effectiveness of primary sedimentation.  
a. True  
*b. False  
2. An inadequate detention time in a primary clarifier would result in increased solids pumping to  
digesters  
a. True  
*b. False  
3. A well-operated primary clarifier, will remove between 90 and 95 percent of the influent settleable  
solids.  
*a. True  
b. False  
4. Baffles ahead of outlet weirs in primary tanks serve as a physical barrier to inhibit the loss of  
floating solids and grease.  
*a. True  
b. False  
5. Baffles upstream of outlet weirs in primary tanks serve as a physical barrier to inhibit the loss of  
floating solids and grease.  
*a. True  
b. False  
6. Detention time is the time required to fill a tank at a given flow.  
*a. True  
b. False  
7. Gas bubbles rising to the surface in a primary clarifier are normally an indication that the sludge in  
the unit is undergoing proper settling  
a. True  
*b. False  
112  
8. Hydraulic loading to a clarifier is expressed as gallons/day.  
Chapter 7. Primary Treatment  
a. True  
*b. False  
9. It is common for primary clarifiers to remove 80-85% of the BODand90-95% of the TSS.  
a. True  
*b. False  
10. Operators may use sludge blanket levels in primary clarifiers to adjust sludge pumping rates.  
*a. True  
b. False  
11. The weir overflow rate of a primary clarifier is expressed as gal/day/ft2 (gallons per day per square  
feet)  
a. True  
*b. False  
12. Primary treatment is a process which allows substances that will settle or float to be separated from  
the wastewater being treated.  
*a. True  
b. False  
13. A well operating primary clarifier will generally remove nearly all of the influent suspened solids.  
a. True  
*b. False  
14. Sludges from primary clarifiers are usual ly more dense than sludges from secondary clarifiers.  
*a. True  
b. False  
15. The generally accepted range of BOD removal in a well operating primary clarifier would be 40 -  
60% .  
a. True  
*b. False  
16. Wooden flights in a rectangular clarifier usually are equipped with wearing shoes.  
*a. True  
b. False  
17. The typical percentage of BOD which should be removed in primary clarifiers is in the range of  
about 90%-99%  
a. True  
*b. False  
18. If the collector mechanism on a circular clarifier stalls, it would be advisable to reverse the motor  
and back up the collector.  
a. True  
*b. False  
19. Raw sludge drawn from a primary sedimentation tank normally would contain 1,000 mg/l - 3,000  
mg/1 of settleable solids.  
a. True  
*b. False  
20. When pumping primary clarifier scum to a holding tank, there should be no additional problems in  
cold weather.  
a. True  
7.4 Advance primary treatment (APT)  
113  
*b. False  
21. If a shear pin for a particular installation is constantly bending, it would be proper and advisable to  
replace it with a pin of greater shear strength.  
a. True  
*b. False  
22. Septic sludge generally has a low pH.  
*a. True  
b. False  
23. Short circuiting in a primary clarifier occurs mainly because of the effluent weirs being plugged or  
not level  
*a. True  
b. False  
24. Wear shoes are found in circular clarifiers  
a. True  
*b. False  
25. Hydraulic loading is the number of gallons flowing each day through one cubic foot volume of the  
clarifier  
a. True  
*b. False  
26. Clarifier detention time is the time required to fill the clarifier at a given flow rate  
*a. True  
b. False  
27. Main objective of primary treatment is to remove organics present in wastewater  
a. True  
*b. False  
28. Raw sludge pumped from a primary clarifier generally contains about 3% to 6% total solids.  
*a. True  
b. False  
29. The main function of launders in a rectangular sedimentation tank is to prevent scum and other  
floatables from leaving with the primary effluent.  
*a. True  
b. False  
30. The solids in primary sludge contain both volatile and fixed parts.  
*a. True  
b. False  
31. Uneven weirs (heights) may result In short circuiting in a primary clarifier  
*a. True  
b. False  
32. “V” -notch weirs attached to a primary effluent launder measure the flow rate.  
*a. True  
b. False  
33. A well-operated primary clarifier, will remove between 90 and 95 percent of the influent settleable  
solids.  
*a. True  
b. False  
114  
Chapter 7. Primary Treatment  
34. A well operating primary clarifier will generally remove nearly all of the influent settleable solids.  
*a. True  
b. False  
35. In a well-designed, efficiently operated wastewater treatment facility, most of the colloidal and  
dissolved matter in wastewater will be removed in the primary clarifiers  
a. True  
*b. False  
36. In some primary clarifiers water or aIr sprays may be used to push scum toward the scum removal  
point.  
*a. True  
b. False  
37. The main purpose of primary treatment is to remove the materials that might damage downstream  
equipment and/or the pumps.  
a. True  
*b. False  
38. The volume of sludge to be removed from a primary settling tank maybe estimated by measuring  
the primary influent and effluent settleable solids and wastewater flow.  
*a. True  
b. False  
39. A device called an Imhoff cone is commonly used to measure settleable solids in:  
a. Percent  
*b. mL/L  
c. mg/L  
d. ppm  
e. SVI units  
40. An efficient primary clarifier is expected to remove what percent of the influent settleable solids?  
a. 2 to 6 %  
b. 20 to 40 %  
c. 40 to 50 %  
d. 60 to 75 %  
*e. 90 % or better  
41. A primary clarifier does not have adequate detention time. Which of the following would result?  
a. Decreased organic leading on secondary unit  
b. Overloading of collector or flight drive motor  
c. Increased solids pumped to digester  
*d. Low BOD removal  
e. Reduced suspended solids in aeration tank.  
42. If the sludge depth in a secondary sedimentation tank is too high, what will happen?  
a. Decreased turbidity in effluent.  
b. Return activated sludge will have lower oxygen demand  
c. Settleable solids from aeration tank will increase  
*d. Sludge may become septic  
43. Odors associated with septic sludge are usually caused by  
a. A neutral pH  
b. Inorganic matter  
7.4 Advance primary treatment (APT)  
115  
c. Short retention time in collection system  
d. Active aerobic bacteria  
*e. Active anaerobic bacteria  
44. The main objectives of primary sedimentation are to remove:  
a. Finely divided particles and dissolved organics  
b. TDS and colloidal solids  
c. BOD, COD, and SS  
*d. Settleable solids and floatables  
e. Detritus and volatile organics  
45. The withdrawal of sludge from a primary clarifier should be slow in order to:  
a. conserve electricity  
*b. prevent pulling too much water with the sludge  
c. keep the BOD stabilized  
d. not disturb the bacteria in the clarifier  
e. protect the pump  
46. The withdrawal of sludge from a clarifier should be slow in order to:  
a. protect the pump.  
b. conserve electricity.  
*c. prevent the pulling of too much water with the sludge.  
d. not disturb the bacteria in the digester.  
e. avoid breaking the water seal in the digester.  
47. Which of the following terms refers to a hydraulic condition, typically indicated by billowing solids  
flowing over the effluent weir, where a portion of the flow through a clarifier experiences a much  
shorter detention time than the rest of the wastewater in the tank?  
a. surging  
*b. shortcircuiting  
c. overload  
d. dispersion  
48. When wastewater enters a rectangular clarifier, it is often evenly dispersed across the entire width  
of the tank by means of:  
*a. a baffle.  
b. a proportional weir.  
c. a submerged launder.  
d. a dome diffuser.  
e. a hydraulic pump.  
49. Rectangular primary clarifiers have wooden or plastic flights, which are equipped with:  
a. shear pins.  
b. friction skirts.  
*c. wear shoes.  
d. grease nipples.  
e. sludge hinges.  
50. The main function of an inlet baffle in a settling tank is to:  
*a. Reduce velocity and disperse the flow  
b. Increase velocity to prevent excessive settling near the inlet  
c. Remove scum from the wastewater  
116  
Chapter 7. Primary Treatment  
d. Protect the scrapping mechanism from damage by excessive velocities.  
51. If short circuiting occurs in a clarifier, the operator should  
*a. Identify the cause  
b. Change fuses.  
c. Increase the sludge drawoff  
d. Restart the pump.  
52. The expected range of BOD removal in a well operated primary clarifier is:  
a. 10 to 20%  
*b. 25 to 40%  
c. 40 to 60%  
d. 60 to 80%  
53. A treatment plant process using sedimentation after screening and grit removal and before aerobic  
treatment is known as:  
a. preliminary treatment.  
*b. primary treatment.  
c. secondary treatment.  
d. RBC treatment.  
e. advanced wastewater treatment.  
54. Sludge gasification in a primary clarifier may be the result of:  
a. hydraulic overloading  
b. low influent BOD concentrations  
*c. infrequent pumping of sludge  
d. too short of detention time  
55. The withdrawal of sludge from a primary clarifier should be slow in order to:  
a. conserve electricity  
*b. prevent pulling too much water with the sludge  
c. keep the BOD stabilized  
d. not disturb the bacteria in the clarifier  
e. protect the pump  
56. Primary sedimentation will remove most of the:  
a. BOD and suspended solids.  
b. settleable solids and suspended solids.  
c. grit and raw sludge.  
*d. floatable and settleable solids.  
e. suspended solids and pathogens.  
57. The time it takes for a unit volume of wastewater to pass entirely through a primary clarifier is  
called:  
*a. detention time  
b. hydraulic loading rate  
c. overflow time  
d. weir loading rate  
58. Primary treatment units are designed to remove settleable solids from the wastewater stream by:  
a. biological treatment  
b. chemical addition  
c. biofiltration  
7.4 Advance primary treatment (APT)  
117  
*d. gravity sedimentation  
e. comminuting devices  
59. Which one of the following factors would have least influence on the settleability of solids in a  
clarifier:  
a. detention time  
b. flow velocity  
c. short-circuiting  
d. temperature  
*e. soluble BOD  
60. The least acceptable method of handling scum and skimmings from a primary clarifier is:  
a. burning  
b. burial  
c. pumping to a digester  
*d. discharging into the effluent  
e. hauling to a sanitary landfill  
61. Given the following data, what is the most likely cause of the primary sedimentation tank problem?  
DATA: Raw sludge pumps run 10 minutes in each hour Raw sludge has 3% total solids at start  
of pumping cycle, 2% total solids at end. No sludge accumulation on tank floor Slight sludge  
accumulation in sludge hopper  
a. Raw sludge pumping duration too short  
*b. Raw sludge pumping duration too long  
c. Raw sludge total solids too high  
d. Sludge collectors running too long  
62. Raw sludge pumping cycles that are too short will  
*a. Cause wastestream going to secondary treatment to turn dark.  
b. Decrease return and waste sludge suspended solids.  
c. Increase amount of grit in raw sludge.  
d. Increase rags in raw sludge.  
63. Raw sludge should be removed from a primary settling tank at any plant  
a. At least hourly.  
b. Not more often than once a week.  
*c. At least once a day.  
d. Whenever sludge rises to the surface.  
118  
Chapter 7. Primary Treatment  
64. You are the operations supervisor at a modern 27 MGD (average dry weather flow) conventional  
activated sludge wastewater treatment plant. During a recent plant expansion, four (4) new primary  
sedimentation tanks (160 ft. long x 25 ft. wide x 12 ft. deep) were built. These tanks have been  
tested and are ready for service. These new tanks are to be put on line while six old sedimentation  
tanks are to be taken out of service so that they may be inspected and repaired. The primary influ-  
ent channel and other mechanical equipment (e.g. pumps, collector mechanisms, flights, etc.) will  
also be inspected and repaired. It is estimated that the total down time will be one month or more.  
Because of the design of the primary influent channel, all six of the old sedimentation tanks must  
be taken out of service at one time.  
The table below summarizes the calculated surface loading rates and detention times with the  
original and when only four tanks are on line.  
FLOW CONDITION  
Ave. Daily Flow  
Peak Flow  
SURFACE LOADING  
1688 gpd/ ft2  
DETENTION TIME  
1.3 hours  
2531 gpd/ ft2  
0.85 hours  
Note: Peak flow is 1.5 times the average daily flow.  
Write a memo to your shift supervisors in which you identify and briefly discuss FIVE (5) signifi-  
cant concerns in regard to a plan of action (POA) for removal and inspection of the old sedimenta-  
tion tanks. State all assumptions.  
7.4 Advance primary treatment (APT)  
119  
Response:  
Assessing the clarifier surface loading rate for the new clarifiers:  
Surface area of new tanks= 160ft 25 ft = 4,000ft2/tank = 16,000ft2 total surface area - four  
new clarifiers  
Clarifier surface loading - average dry flow condition:  
6
Clari fier influent flow(gpd)  
Clari fier sur face area(ft )  
gpd  
ft  
2710 gpd  
gpd  
ft  
=
=⇒  
= 1,688  
2
2
2
2
16,000ft  
Assessing the clarifier detention time for the new clarifiers:  
Volume of new tanks= 160 ft 25 ft 12 ft = 48,000 ft3/tank = 288,000ft3 total volume - four  
new clarifiers  
Clarifier detention time - average dry flow condition:  
3
3
Clari fier volume(ft )  
288,000ft  
(hrs) =  
=⇒  
= 1.91 hrs  
3
3
ft  
ft  
gal  
day  
6
flow(  
)
2710  
hr  
day 24hrs 7.48gal  
Both, detention time & the surface loading rate of the new clarifiers is comparable to the existing  
ones - so no concern from that perspective.  
i Ensure the effluent weirs in the new clarifiers are level to ensure there is no short circuiting  
ii Ensure that the new clarifier mechanical equipment and its associated control systems are  
tested to ensure reliable operations prior to the switch.  
iii For the inspection/repair of the old tanks, ensure lockout-tagout and proper confined entry  
procedures are followed  
120  
Chapter 7. Primary Treatment  
65. Chemical enhanced primary treatment (CEPT) is used at a number of conventional activated sludge  
planes. CEPT, also called advanced primary treatment, typically involves the application of liquid  
ferric chloride and an anion polymer ahead of the primary sedimentation tank. Although the long-  
term addition of chemicals to a wastewater flow can be costly, the benefits of CEPT may outweigh  
its cost.  
Answer the fol- lowing question about advanced primary treatment:  
(a) Identify two operational considerations in order optimize CEPT.  
(b) Identify and explain two ways in which CEPT could result in energy saving in a convention  
activated sludge plant that uses cogeneration  
(c) Identify two additional benefits to anaerobic digester operations that may result from the  
addition of ferric chloride. Briefly explain how ferric chloride is able to yield these benefits.  
Response:  
a Identify two operational considerations in order optimize CEPT.  
• Optimal ferric chloride and polymer dosage  
• Adequate mixing time and energy for ensuring proper coagulation by ferric chloride  
Ensure the polymer is gently folded into the coagulated wastewater just prior to entering  
the clarifier.  
b
Identify and explain two ways in which CEPT could result in energy saving in a convention  
activated sludge plant that uses cogeneration.  
By increasing BOD removal in the primary clarifier, the influent BOD loading to the  
secondary treatment is lower reducing the oxygen requirements thus resulting in energy  
savings.  
The organic material in the sludge feed to the digester has a higher proportion of pri-  
mary sludge which is more easily digestable thus producing more digester gas for cogen-  
eration.  
c
Identify two additional benefits to anaerobic digester operations that may result from the  
addition of ferric chloride. Briefly explain how ferric chloride is able to yield these benefits.  
The residual ferric chloride in the primary sludge feed allows for the control of H2S  
concentration in the digester gas  
Ferric chloride also helps in minimizing struvite forming potential in the digester due to  
its ability to remove phosphate, one of the key elements of struvite.  
7.4 Advance primary treatment (APT)  
121  
66. Recently your activated sludge plant has acquired a two 2-meter belt press to dewater anaerobically  
digester sludge.  
(a) List four (4) factors, other than belt speed, which generally affect the belt press performance.  
(b) The belt press is running. You observe what one of your operator’s calls washout. Define  
"washout."  
(c) Identify two possible cause of washout.  
(d) It has been said that the "the ideal belt speed is the slowest the operator can maintain without  
causing washout." Explain why this is so.  
Response:  
a List four (4) factors, other than belt speed, which generally affect the belt press performance.  
Primary sludge dewaters better than secondary sludge. Digested sludge with a higher  
proportion of primary will produce drier (higher solids content) dewatered biosolids.  
• Optimal dosage and adequate mixing of polymer with the sludge.  
• Cleanliness of the belts.  
• Adequate and even hydraulic pressure to ensure water is squeezed out from the sludge.  
b Identify two possible cause of washout.  
• Inadequate polymer dosing.  
• Hydraulic overloading.  
• Blinded gravity zone belt.  
• Slow belt speed  
c
It has been said that the "the ideal belt speed is the slowest the operator can maintain with-  
out causing washout." Explain why this is so.  
As the belt speed is slowed, the water from the sludge will have more time to drain producing  
a drier cake. However, due to the slow belt speed, solids will build up on the belt preventing  
the water from draining. Washout occurs when the belt speed is slowed to a point when there  
is so much standing water that it flows over the sides of the belt.  
8. Introduction to Secondary Treatment  
While preliminary and primary treatment processes are designed primarily to remove solids from  
wastewater, secondary treatment is for the removal of organics.  
• Secondary treatment involves:  
biological conversion of the dissolved and suspended organics in wastewater into biomass,  
and  
physical settling (separation) process where the solids including the biomass formed during  
secondary treatment is separated and removed from the treated wastewater.  
With the removal of gross solids in the preliminary treatment followed by removable of settleable  
solids in the primary clarifiers and the removal of dissolved and suspended organics in the sec-  
ondary treatment processes, the wastewater is considered treated.  
Secondary treated wastewater is typically disposed or treated further for reuse or disposal (depend-  
ing upon the end use/application and the NPDES permit stipulations).  
The solids (biomass) removed from the secondary treatment is typically mixed with the solids from  
primary treatment and stabilized using a solids treatment process like sludge digestion prior to its  
disposal.  
Secondary treatment process incorporates one of the following three approaches:  
8.1 Fixed film system  
Here the microorganisms responsible for the treatment, grow on substrates such as rocks, sand or  
plastic.  
When the wastewater is spread over the substrate, the microorganisms up-take the organics present  
in the wastewater  
Example of this secondary treatment process include trickling filters and rotating biological contac-  
 
 
124  
Chapter 8. Introduction to Secondary Treatment  
tors  
8.2 Suspended Growth System  
In this type of secondary treatment, the microbes are suspended in the wastewater flow being  
treated.  
Air or oxygen is supplied to maintain an aerobic environment and to keep the microorganisms in  
suspension.  
• Example of this secondary treatment approach include the activated sludge treatment process  
8.3 Pond System  
Similar to the suspended growth, stabilization ponds are large man made bodies of water which treat  
wastewater using mainly natural processes including sunlight, algae and microorganisms.  
 
 
9. Trickling Filters  
9.1 Theoretical Background  
Trickling filter is a fixed film secondary treatment process wherein the organic content of the wastewater is  
removed using biological growth attached to an inert media such as lava rock or plastic  
In a trickling filter, the wastewater is sprayed evenly on the surface of the media with a rotary type  
distributor with orifices  
The wastewater percolates through the media bed, where it comes in contact with biological slime  
growth – zoogleal film (zooglea)  
The aerobic biomass - bacteria, protozoa and other microoorganisms in the zooglea capture and  
consume the suspended and dissolved organics from the wastewater.  
The microorganisms metabolize the organics and in the process produce more microbial mass  
resulting in increasing the thickness of the zoogleal layer.  
The thickness of the zoogleal layer can only increase to a point until the wastewater flow – hy-  
draulic load, shears the slime layer – “sloughs off” and is carried out as part of the effluent flow as  
sloughing.  
The treated wastewater cascades from the bottom of the media into the underdrain system – lower  
portion of the TF comprised of columns which support the media base. The underdrain has a  
sloping floor to direct the cascading water into a center channel .  
The clarifier allows for the separation (settling) of the of the solids (sloughed off material). The  
settled solids is removed - typically pumped to a digester and the clarified effluent flows out of the  
clarifier.  
The source of oxygen to support the aerobic growth is from the oxygen dissolved in the wastewater  
as it is sprayed over the media and from the air currents due to the downward flow of the wastew-  
ater and the temperature difference between ambient and the interior of the trickling filter. Forced  
ventilation system may be designed as part of the trickling filter  
• Word trickling “filter” is a misnomer - no filtration is involved  
 
 
126  
Chapter 9. Trickling Filters  
• Advantage includes process simplicity and lower costs  
• Disadvantage include BOD removal efficiency of only about 80-85  
The media may be rock, slag, coal, bricks, redwood blocks, molded plastic, or any other sound  
durable material.  
The media depth ranges from about three to eight feet for rock media trickling filters and 15 to 30  
feet for synthetic media.  
The media needs to be uniformly sized and have adequate empty spaces (voids) to ensure maintain-  
ing aerobic condition necessary for the survival of biomass.  
Pre-fabricated (synthetic) media - similar to the one shown below, has an advantage over the  
"dumped" type media such as lava rock of providing a greater surface area per volume upon which  
the zoologeal film may grow while providing ample void space for the free circulation of air.  
Sometimes, due to inadequate hydraulic loading, portions of the zoogleal layer may become too  
thick and oxygen cannot penetrate its full depth, causing odor issues.  
9.2 Trickling Filter Recirculation  
Recirculation - where a portion of the treated wastewater is returned back as the feed to the TF. The  
parameter recirculation ratio is calculated to quantify the recirculation flow. Recirculation ratio is a ratio  
of the recirculated flow - QR to the influent flow QI. Recirculation ratios typically vary from 0.5 to 4.  
Recirculation is beneficial for the following reasons:  
It improves the removal efficiency by increasing the contact time of the zoogleal layer with the  
wastewater  
• During low flows it prevents the trickling filter from drying out  
• It dilutes any toxic loadings  
• It promotes oxygen transfer and reduces ponding  
The increased hydraulic loading promotes uniform sloughing, prevents ponding, improves ventila-  
tion through the filter and reduces potential for snail and filter fly breeding  
 
9.3 Operation of Multiple Trickling Filters  
127  
9.3 Operation of Multiple Trickling Filters  
Multiple trickling filters can be operated in series or in parallel:  
• Series operation in which the flow from one flows into the next.  
For high strength loading and for nitrification  
• Parallel operation in which the trickling filters that are operated side by side.  
For winter operation - prevents freezing in the TF  
9.4 Parameters for monitoring and operating trickling filters  
1. Hydraulic loading is expressed as gpd/ ft2  
2. BOD Removal (%)  
3. Organic loading lbs BOD/day/ 1000 cu ft  
4. Recirculation ratio  
9.5 Classifying Trickling Filters  
Trickling filters are classified according to the hydraulic and organic loading applied to the filter  
9.5.1 Low-rate filter  
The standard rate or low rate trickling filters (LRTF) are relatively simple treatment units that  
normally produce a consistent effluent quality even with varying influent strength  
• They are generally not provided with recirculation of effluent  
Depending upon the dosing system, wastewater is applied intermittently with rest periods which  
generally do not exceed five minutes at the designed rate of waste flow.  
While there is some unloading or sloughing of solids at all times, the major unloadings usually  
occur several times a year for comparatively short Periods of time.  
Hydraulic loading is 25 - 100 gal/day/sq. ft  
BOD Removal (%) 50 – 80%  
Organic loading is 5 - 25 lbs BOD/day/1000 cu ft  
9.5.2 High-rate filter  
The most important element of a high-rate trickling filter is the provision where a part of the settled  
treated effluent is pumped to the PST or to the filter. This is termed as Recirculation  
 
 
 
 
 
128  
Chapter 9. Trickling Filters  
High-rate filters are usually characterized by higher hydraulic and organic loadings than low-rate  
filters  
The higher BOD loading is accomplished by applying a larger volume of waste per unit surface  
area of the filter.  
As a result of the higher flow velocities a more continuous and uniform sloughing of excess  
zoogleal growth occurs  
Hydraulic loading is 100 - 1000 gal/day/sq. ft  
BOD Removal (%) 65 - 85%  
Organic loading is 25 - 100 lbs BOD/day/ 1000 cu ft  
9.5.3 Roughing filter  
• Roughing filters are high rate type filters designed with plastic packing  
• In most cases roughing filters are used to treat wastewater prior to secondary treatment  
One of the advantages of roughing filter is low energy requirement for BOD removal of high  
strength wastewaters as compared to activated sludge process because the energy required is only  
for pumping the influent wastewater and recirculation flows  
9.6 Trickling Filter Operational Issues  
9.6.1 Ponding  
If the voids in the media get plugged, flow can collect on the surface in ponds. Correction: spraying the  
surface with high pressure water stream stopping a rotary distributor over the ponded area hand-stir the  
media or open the voids dose the filter with chlorine for several hours  
9.6.2 Odors  
• Corrective measures should be taken immediately if foul odors develop  
• presence of foul odors indicates anaerobic conditions are predominant  
• Check the under drain system for obstructions or heavy biological growths  
• increase the recirculation rate to provide more oxygen to the filter bed and increase sloughing  
• keep slime growths off of sidewalks and inside walls of the filter to reduce the odor  
9.6.3 Trickling Filter Flies - Psychoda  
• tiny, gnat-size filter fly, or Psychoda - primary nuisance insect  
• Correction methods include  
Increase recirculation rate  
keep orifice openings clear  
apply insecticides to filter walls  
dose filter with chlorine  
keep weeds and tall grass cut around filter  
9.6.4 Cold weather problems  
• ice can form on the media of the filter  
• Correction methods include:  
decrease recirculation to the filter (influent is usually warmer than recycled flows)  
construct wind screens  
 
 
 
 
 
 
9.6 Trickling Filter Operational Issues  
129  
operate two-stage filters in parallel rather than in series  
Chapter Assessment  
1. Compared to activated sludge, trickling filters are sturdy work units not easily up set by shock  
loads.  
*a. True  
b. False  
2. Ponding on the surface of trickling filters is almost always caused by plugging of the underdrains  
immediately below the point of ponding  
a. True  
*b. False  
3. Zoogleal mass sheared of from the trickling filter media and carried out as part of the effluent flow  
is termed as sloughing  
*a. True  
b. False  
4. For multiple trickling filter operation, the parallel mode is more suitable than the series mode for  
treating influent wastewater with a high BOD content  
a. True  
*b. False  
5. Synthetic /pre-manufactured media used in the biofilter has the advantage of providing a greater  
surface area per volume upon which the zoologeal film may grow while providing ample void  
space for the free circulation of air.  
*a. True  
b. False  
6. Of the two forms of secondary treatment: trickling filters and conventional activated sludge, trick-  
ling filters are more likely to produce a clearer effluent.  
a. True  
*b. False  
7. A trickling filter reduces the strength of wastewater applied to it by the filtering and straining action  
132  
Chapter 9. Trickling Filters  
of the stones or support media.  
a. True  
*b. False  
8. A trickling filter provides higher BOD removal in cold weather than in hot weather.  
a. True  
*b. False  
9. The biological growth on trickling filter media is composed principally of aerobic forms of bacteria,  
fungi, algae, protozoa, worms and the larvae of insects.  
a. True  
*b. False  
10. The spaces between the filter media in a trickling filter need to remain open to permit free flow of  
air throughout the bed.  
*a. True  
b. False  
11. The flooding of filter media with wastewater is sometimes effective in killing filter flies.  
*a. True  
b. False  
12. Sandstone, because of its ability to provide excellent ventilation usually is used for media in trick-  
ling filters.  
a. True  
*b. False  
13. When a trickling filter is referred to as a "roughing filter", it means the operational life of the filter  
is at an end, and a major overhaul is eminent.  
a. True  
*b. False  
14. The hydraulic loading applied. to a trickling filter is the total volume of liquid, including recircula-  
tion.  
*a. True  
b. False  
15. The main function of a launder in a secondary clarifier is to prevent scum and other floatables from  
leaving with the effluent flow  
a. True  
*b. False  
16. The trickling filter recirculation ratio is calculated as Qi/Qr where Qi is the influent flow and Qr is  
the recirculated flow.  
a. True  
*b. False  
17. The advantage of trickling filter over the activated sludge lies in its BOD removal efficiency  
a. True  
*b. False  
18. Recirculation helps prevent excessive sloughing  
a. True  
*b. False  
19. Series operation of trickling filter is the preferred mode during very low temperature conditions  
during winter  
9.6 Trickling Filter Operational Issues  
133  
a. True  
*b. False  
20. A good reason to run a trickling filter in the series mode of operation is:  
a. During cold weather to prevent ice formation  
b. When the influent BOD loading is low  
*c. When the influent BOD loading is high  
d. When the weather is warm  
21. A good reason to run a trickling filter in the parallel mode of operation is:  
a. When the weather is warm  
b. When the BOD loading is high  
*c. When the BOD loading is low  
d. To prevent filter flies  
22. A pan test should be done monthly on a trickling filter to:  
a. Check oil quality in the distributor bearing  
*b. Check sewage distribution on filter  
c. To check flow rates through the media  
d. All the above  
e. None of the above  
23. A problem associated with trickling filters is  
a. Bulking  
b. Protozoans  
*c. Ponding  
d. Liquefaction  
24. A trickling filter or activated sludge process causes nitrification. which is  
a. Conversion of nitrogen to nitrate  
b. Conversion of nitrogen to ammonia  
c. Conversion of nitrate to nitrogen  
*d. Conversion of ammonia to nitrate and nitrite nitrogen  
25. If a trickling filter has been operating with a hydraulic loading (including some recirculation) of  
10 to 12 MGD and organic loading of about 80 pounds of BOD per 1,000 cu. ft./day the treatment  
efficiency will usually increase if the recirculation is increased. This might be attributed to:  
a. The increased recirculation wears down the soluble BOD to finer particles  
*b. The increased flow more completely wets and contacts all of the slime surfaces in the filter so  
the food to effective microorganism ration is less as in an activated sludge process  
c. The grazing population of warm and other organisms in the filter is flushed out before they  
consume the slime bacteria  
d. The increased flow more completely fills the under drains so cold updrafts are eliminated  
e. The statement is just poppycock put out by power companies to get us to use more electricity to  
run pumps  
26. Most common trickling filter operational control method is:  
a. Sloughing control  
*b. Recirculation  
c. Sludge removal  
d. Distributor arm speed  
27. The media in trickling filters is placed  
134  
Chapter 9. Trickling Filters  
a. On a rubber tile floor.  
*b. On a system of tile or columnar underdrains.  
c. Directly in a concrete slab.  
d. Directly in the ground  
28. Sloughing from a trickling filter refers to  
a. A process whereby wastewater is recirculated over the filter.  
b. Bypassing of the filter.  
c. The breaking of the filter stone as a result of weathering and the sluicing of small stone particles  
to the final settling tank.  
*d. The periodic discharge of large quantities of slime growth with the filter effluent.  
29. The primary organisms responsible for treating wastewater in the trickling filter process are:  
a. Anaerobic bacteria  
b. Anoxic bacteria  
c. Facultative bacteria  
*d. Aerobic bacteria  
30. The primary organisms responsible for treating wastewater in the trickling filter process are:  
a. Anaerobic bacteria  
b. Anoxic bacteria  
c. Facultative bacteria  
*d. Aerobic bacteria  
31. Ponding of a trickling filter means  
a. Flooding the filter.  
b. Hosing the growth from the rocks with a high pressure hose.  
*c. Blockage in the media that s prevents water to flow through.  
d. Running the trickling filter effluent to a wastewater pond.  
32. Which of the following would be a cause for trickling filter ponding:  
a. Increased recirculation  
b. Filter flies  
c. Zoogleal mass  
*d. An excessive organic loading without a corresponding high recirculation rate  
33. A good reason to run a trickling filter in the series mode of operation is:  
a. During cold weather to prevent ice formation  
b. When the influent BOD loading is low  
*c. When the influent BOD loading is high  
d. When the weather is warm  
34. A good reason to run a trickling filter in the parallel mode of operation is:  
a. When the weather is warm  
b. When the BOD loading is high  
*c. When the BOD loading is low  
d. To prevent filter flies  
35. A problem associated with trickling filters is  
a. Bulking  
b. Protozoans  
*c. Ponding  
d. Liquefaction  
9.6 Trickling Filter Operational Issues  
135  
36. A trickling filter or activated sludge process causes nitrification. which is  
a. Conversion of nitrogen to nitrate  
b. Conversion of nitrogen to ammonia  
c. Conversion of nitrate to nitrogen  
*d. Conversion of ammonia to nitrate and nitrite nitrogen  
37. Which test best measures the efficiency of a trickling filter?  
a. Total solids.  
b. pH  
*c. BOD  
d. Temperature  
e. Sludge age  
38. What are sloughings?  
a. Troughs.  
b. Slop.  
*c. Zoogleal mass washed off trickling filter media.  
d. Waste-activated sludge.  
e. Grit.  
39. Most common trickling filter operational control method is:  
a. Sloughing control  
*b. Recirculation  
c. Sludge removal  
d. Distributor arm speed  
40. Most common trickling filter operational control method is:  
a. Sloughing control  
*b. Recirculation  
c. Sludge removal  
d. Distributor arm speed  
41. The rotating biological contactors operate based upon the same principles as:  
a. Aerated lagoons  
b. Activated sludge systems  
*c. Trickling filters  
d. Extended aeration  
42. A two-stage trickling filtration system means:  
a. one filter used with recirculation returning from filter effluent to filter influent at 100b. one filter  
used with recirculation from secondary clarifier effluent to trickling filter influent  
*c. two filters used in series, either directly or with a clarifier in between  
d. two filters used in parallel  
e. one filter used with no primary clarification  
43. Regardless of shape, one acre-foot of media in a trickling filter is equal to:  
a. 33,000 cu. ft.  
*b. 43,560 cu. ft.  
c. 55,560 cu. ft.  
d. 77,840 cu. ft.  
e. none of the above  
44. The presence of a "rotten egg" odor in the area of a trickling filter generally would indicate which  
136  
Chapter 9. Trickling Filters  
of the following:  
a. the presence of the psychoda fly  
b. the clogging of the distributor arm orifices  
*c. anaerobic conditions within the filter  
d. too much recirculation  
e. too high a DO in the: wastewater being applied to the filter  
45. Analysis of a wastewater effluent from a standard rate trickling filter shows the following:  
Nitrite= 8 mg/L  
Ammonia= 2 mg/L  
Nitrate= 2 mg/L  
What do these results indicate?  
*a. Plant is operating properly  
b. Final end product, nitrite, is too low  
c. Final end product, nitrate, is too low  
d. DO was insufficient to allow complete aerobic oxidation of the nitrate to nitrite  
46. Given the data below, what is the most likely cause of the trickling filter problem? –  
DATA: Normal dry weather plant flow rate  
Sweep arms operating at normal rate  
Water not passing through as rapidly as normal  
Increased odor level  
Increased chlorine demand in final effluenTrue  
Final effluent turbid.  
*a. Filter media plugged with dead microorganisms  
b. Low solids load to filter  
c. seal in sweep arms worn  
d. Spray nozzles partially plugged.  
47. The rock/media in most trickling filters is placed  
a. Directly in the ground.  
b. Directly in a concrete slab.  
*c. On a system of tile underdrains  
d. On a rubber tile floor  
48. The primary organisms responsible for treating wastewater in the trickling filter process are:  
a. Anaerobic bacteria  
b. Anoxic bacteria  
c. Facultative bacteria  
*d. Aerobic bacteria  
49. Trickling filters usually consist of a bed of stone which performs its function in sewage by:  
a. Mechanical filtration of organic solids  
b. Mechanical filtration of inorganic solids  
c. Provides aeration  
*d. Supports a growth of organisms which feed upon the sewage and reduce BOD  
e. None of the above  
50. What are sloughings?  
a. Troughs.  
b. Slop.  
9.6 Trickling Filter Operational Issues  
137  
*c. Zoogleal mass washed off trickling filter media.  
d. Waste-activated sludge.  
e. Grit.  
51. Which of the following is not a benefit of recirculation in the trickling filter process?  
a. Dilutes high strength or toxic wastes.  
b. Helps prevent septic conditions in trickling filter  
*c. Helps prevent excessive sloughing  
d. Helps control odors, ponding and filter flies  
10. Stabilzation Ponds  
Stabilization ponds and lagoons are bodies of water which treat wastewater using mainly natural  
processes including sunlight, algae and microorganisms for treating wastewater  
• While ponds are shallow and man-made, lagoons are bodies of water confined within natural bound-  
aries.  
10.1 Advantages of ponds  
• Cheap to build and operate  
• Low maintenance and electrical costs  
• Do not require highly trained operational personnel  
Provide treatment that can be equal to some secondary treatment processes and have fewer sludge  
handling issues.  
10.2 Disadvantages of ponds  
• Land intensive  
• Effluent quality varies with seasonal temperature changes  
• Suspended solids levels that can create regulatory problems.  
• System upsets almost always result in odor problems and recovery times may be weeks or months.  
• Not appropriate for colder climates  
10.3 Types of Ponds  
10.3.1 Anaerobic Ponds  
Anaerobic Ponds  
• Typically for treating raw sewage  
 
 
 
 
 
140  
Chapter 10. Stabilzation Ponds  
These are deep - 10-14 feet treatment ponds which rely primarily on anaerobic bacteria to break  
down the organic waste.  
• Designed for BOD removal.  
• High strength wastewater may be treated.  
Organic matter is broken down releasing releasing methane, carbon dioxide and odorous gases  
including hydrogen sulfide.  
• Most of the decomposition is accomplished by acid forming bacteria.  
• The pH in these lagoons is usually below 6.5.  
• They are total retention and do not have an effluent discharge.  
• The anaerobic pond must be de-sludged approximately once every 2 to 5 years  
• Organic loading of 200-1000 lbs. BOD5 per acre per day  
10.3.2 Facultative Ponds  
The depth of facultative ponds is about 4-7 feet which is in-between the depths of anaerobic ponds  
(10-14 feet) and aerobic ponds 3 feet)  
• The uper layer of facultative pond is aerobic, and bottom layer is mostly anaerobic.  
Facultative bacteria are responsible for most of the treatment that occurs in these ponds. Facultative  
bacteria are bacteria which can live under both aerobic and anaerobic conditions.  
• The algae that grow in the pond are critical to the successful stabilization of the organic load.  
The algae will take in carbon dioxide (CO2) and, through photosynthesis, use it to create sugars and  
release dissolved oxygen (O2) that is used by the aerobic bacteria. Facultative lagoon levels should  
always maintain at least 4 feet of water in the pond.  
• Typically for secondary treatment - BOD removal  
• 15-50 lbs BOD5 per acre per day.  
Unused CO2 will react with water to form carbonic acid - which would reduce the pH unless con-  
sumed  
Sludge removal need is rare. Sludge can be removed by using a raft-mounted sludge pump or by  
draining and dewatering the pond and removing the sludge with a front-end loader.  
10.3.3 Aerobic Stabilization Ponds  
Aerobic stabilization ponds are also known as: maturation, polishing or finishing Pond  
• Contain disssolved oxygen throughout entire depth of the pond.  
Treatment is accomplished through the stabilization of organic wastes by aerobic bacteria and  
algae.  
• Typically for tertiary treatment  
• Designed for pathogen removal  
• Shallow - only about 3 feet deep.  
• They are most often the final cells in a multi-staged pond system  
• They are also used as polishing ponds for tertiary treatment of trickling filter plant effluent.  
• Usually the effluent is directed into a second pond where the sludge can settle  
Their shallow depth allows sunlight to penetrate to the bottom of the pond to encourage algae  
growth and aerobic conditions throughout the pond  
The low solids loading found in these tertiary treatment applications means that these ponds nor-  
mally have no sludge zone  
• These ponds may be mechanically aerated  
 
 
10.3 Types of Ponds  
141  
142  
Chapter 10. Stabilzation Ponds  
• Aerobic polishing ponds are designed for 15-20 pounds BOD/acre/day  
• Aerobic ponds are typically designed for pathogen removal  
• Aerobic lagoon levels should always maintain at least 18 inches of water in the pond  
10.4 Ponds Operations and Maintenance  
• Ponds and lagoons are designed as continuous discharge, controlled discharge, or no discharge.  
In controlled discharge ponds the wastewater is held for long periods of time before discharg-  
ing.  
No-discharge ponds the inflow rate needs to be equal or exceed the rates of evaporation  
and/or percolation.  
Short-circuiting in a pond may be caused by poor design of inlet and outlet piping arrangements or  
by uncontrolled growth of water weeds.  
Stagnant water will breed mosquitoes and can result in anaerobic conditions developing that can  
cause odor issues  
• Dikes need to be maintained  
• Aquatic plants and weeds must be removed from the water.  
• Reeds will create stagnant areas along the edge of the pond and need to be removed  
• To start a new pond, two feet of water is typically added prior to fresh starting wastewater feed.  
Sodium nitrate can be used to help recover from an odor-causing upset. The nitrates (NO3) will  
provide a source of chemically bound oxygen for the bacteria to use instead of dissolved oxygen.  
• Scum control may be required  
 
Chapter Assessment  
1. Organic loading to a stabilization pond is always calculated by knowing the pounds of BOD ap-  
plied per 1.000 cubic feet of pond volume per day.  
a. True  
*b. False  
2. An abundance of aquatic plant growth in and around the edge of a facultative pond provides more  
surface area for biologic activity and increases treatment capacity.  
a. True  
*b. False  
3. Facultative ponds are anaerobic on the bottom and aerobic near the surface.  
*a. True  
b. False  
4. The best time of the year to initiate operation of a new facultative pond is during the coldest months  
of the year.  
a. True  
*b. False  
5. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.  
*a. True  
b. False  
6. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen  
than the surface layer.  
a. True  
*b. False  
7. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-  
lems and to take advantage of the increased solubility of oxygen in cold water.  
a. True  
*b. False  
144  
Chapter 10. Stabilzation Ponds  
8. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million  
gallons.  
*a. True  
b. False  
9. Algae is primarily responsible for the effective working of an anaerobic pond  
a. True  
*b. False  
10. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream  
a. True  
*b. False  
11. Facultative ponds are anaerobic on the bottom and aerobic near the surface.  
*a. True  
b. False  
12. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved  
oxygen.  
*a. True  
b. False  
13. Algae is primarily responsible for the effective working of an anaerobic pond  
a. True  
*b. False  
14. Algae produce oxygen in a facultative pond.  
*a. True  
b.False  
15. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million  
gallons.  
*a. True  
b. False  
16. In a facultative pond, aerobic bacteria produce oxygen that is consumed by algae.  
a. True  
*b. False  
17. In a facultative pond, the pH falls when carbon dioxide produced by the aerobic bacteria is con-  
sumed by the algae  
a. True  
*b. False  
18. In a facultative pond, the algae utilizes the oxygen given off as a by-product of bacterial respiration  
a. True  
*b. False  
19. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen  
than the surface layer.  
a. True  
*b. False  
20. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved  
oxygen.  
*a. True  
b. False  
10.4 Ponds Operations and Maintenance  
145  
21. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen  
than the surface layer.  
a. True  
*b. False  
22. In a facultative wastewater treatment pond that Is operating normally, a rise In DO will be accompa-  
nied by a drop in pH.  
a. True  
*b. False  
23. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-  
lems and to take advantage of the increased solubility of oxygen in cold water.  
a. True  
*b. False  
24. Operational depth of a wastewater pond is not used for calculating its organic loading  
*a. True  
b. False  
25. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.  
*a. True  
b. False  
26. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen  
than the surface layer.  
a. True  
*b. False  
27. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-  
lems and to take advantage of the increased solubility of oxygen in cold water.  
a. True  
*b. False  
28. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million  
gallons.  
*a. True  
b. False  
29. Algae is primarily responsible for the effective working of an anaerobic pond  
a. True  
*b. False  
30. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream  
a. True  
*b. False  
31. The best time of the year to initiate operation of a new facultative pond is during the coldest months  
of the year.  
a. True  
*b. False  
32. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.  
*a. True  
b. False  
33. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen  
than the surface layer.  
146  
Chapter 10. Stabilzation Ponds  
a. True  
*b. False  
34. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-  
lems and to take advantage of the increased solubility of oxygen in cold water.  
a. True  
*b. False  
35. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million  
gallons.  
*a. True  
b. False  
36. A wastewater pond may have a detention time of 30 to 120 days.  
*a. True  
b. False  
37. Sludge deposits are usually in evidence at the effluent end of a waste treatment pond rather than the  
influent end.  
a. True  
*b. False  
38. It is good practice to start waste treatment ponds during winter because of the lower level of biolog-  
ical activity.  
a. True  
*b. False  
39. Anaerobic ponds require higher. minimum temperatures than aerobic ponds to achieve satisfactory  
treatment,  
*a. True  
b. False  
40. The formula for calculating percent BOD removal in pond systems is: BOD removal,  
*a. True  
b. False  
41. The organic loading on a wastewater lagoon is higher than on a polishing pond.  
*a. True  
b. False  
42. Algae produce oxygen in a facultative pond.  
*a. True  
b. False  
43. A disadvantage of excessive scum on an oxidation pond is that it will interfere with photosynthesis.  
*a. True  
b. False  
44. Ranges of treatment efficiencies that can be expected from ponds vary more than most other treat-  
ment processes.  
*a. True  
b. False  
45. At night, algae in a facultative pond consume oxygen.  
*a. True  
b. False  
46. In wastewater treatment ponds, DO levels very rarely exceed 15 mg/l at the pond surface.  
10.4 Ponds Operations and Maintenance  
147  
a. True  
*b. False  
47. Organic loading to a wastewater treatment pond is expressed in units of pounds of VSS per acre per  
day.  
a. True  
*b. False  
48. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream  
a. True  
*b. False  
49. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.  
*a. True  
b. False  
50. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.  
*a. True  
b. False  
51. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved  
oxygen.  
*a. True  
b. False  
52. Algae is primarily responsible for the effective working of an anaerobic pond  
a. True  
*b. False  
53. Algae is primarily responsible for the effective working of an anaerobic pond  
a. True  
*b. False  
54. Algae produce oxygen in a facultative pond.  
*a. True  
b.False  
55. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million  
gallons.  
*a. True  
b. False  
56. In a facultative pond, aerobic bacteria produce oxygen that is consumed by algae.  
a. True  
*b. False  
57. In a facultative pond, the pH falls when carbon dioxide produced by the aerobic bacteria is con-  
sumed by the algae  
a. True  
*b. False  
58. In a facultative pond, the algae utilizes the oxygen given off as a by-product of bacterial respiration  
a. True  
*b. False  
59. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen  
than the surface layer.  
a. True  
148  
Chapter 10. Stabilzation Ponds  
*b. False  
60. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved  
oxygen.  
*a. True  
b. False  
61. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen  
than the surface layer.  
a. True  
*b. False  
62. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved  
oxygen.  
a. True  
*b. False  
63. In a facultative wastewater treatment pond that Is operating normally, a rise In DO will be accompa-  
nied by a drop in pH.  
a. True  
*b. False  
64. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-  
lems and to take advantage of the increased solubility of oxygen in cold water.  
a. True  
*b. False  
65. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-  
lems and to take advantage of the increased solubility of oxygen in cold water.  
a. True  
*b. False  
66. Operational depth of a wastewater pond is not used for calculating its organic loading  
*a. True  
b. False  
67. The best time of the year to initiate operation of a new facultative pond is during the coldest months  
of the year.  
a. True  
*b. False  
68. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.  
*a. True  
b. False  
69. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen  
than the surface layer.  
a. True  
*b. False  
70. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-  
lems and to take advantage of the increased solubility of oxygen in cold water.  
a. True  
*b. False  
71. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million  
gallons.  
10.4 Ponds Operations and Maintenance  
149  
*a. True  
b. False  
72. Algae is primarily responsible for the effective working of an anaerobic pond  
a. True  
*b. False  
73. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream  
a. True  
*b. False  
74. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.  
*a. True  
b. False  
75. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.  
*a. True  
b. False  
76. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved  
oxygen.  
*a. True  
b. False  
77. Ranges of treatment efficiencies that can be expected from ponds vary more than most other treat-  
ment processes.  
*a. True  
b. False  
78. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream  
a. True  
*b. False  
79. The best time of the year to initiate operation of a new facultative pond is during the coldest months  
of the year.  
a. True  
*b. False  
80. The color of the algae at the surface of a facultative pond can be an indication of its operational  
condition.  
*a. True  
b. False  
81. The flow to a wastewater treatment pond may be expressed in units of acre -feet.  
a. True  
*b. False  
82. Usually facultative ponds are mixed with mechanical aerators at night and on cloudy days to dis-  
perse food and bacteria evenly.  
a. True  
*b. False  
83. Wastewater treatment ponds that receive untreated raw wastewater are called stabilization ponds.  
*a. True  
b. False  
84. When a facultative pond develops a green appearance, it should be treated with copper sulfate to  
kill off the algae.  
150  
Chapter 10. Stabilzation Ponds  
a. True  
*b. False  
85. When operating ponds in series, the accumulation of solids in the first pond may become a problem  
after long periods of use.  
*a. True  
b. False  
86. Organic loading to a stabilization pond is always calculated by knowing the pounds of BOD ap-  
plied per 1.000 cubic feet of pond volume per day.  
a. True  
*b. False  
87. An abundance of aquatic plant growth in and around the edge of a facultative pond provides more  
surface area for biologic activity and increases treatment capacity.  
a. True  
*b. False  
88. Facultative ponds are anaerobic on the bottom and aerobic near the surface.  
*a. True  
b. False  
89. The best time of the year to initiate operation of a new facultative pond is during the coldest months  
of the year.  
a. True  
*b. False  
90. Ponds that contain an aerobic top layer and an anaerobic bottom layer are called facultative ponds.  
*a. True  
b. False  
91. In a facultative pond, the bottom layer of the pond will generally contain more dissolved oxygen  
than the surface layer.  
a. True  
*b. False  
92. It is recommended to start up a pond in the coldest months of the year in order to avoid odor prob-  
lems and to take advantage of the increased solubility of oxygen in cold water.  
a. True  
*b. False  
93. A one acre wastewater treatment pond operated at a depth of 3 feet holds approximately 1 million  
gallons.  
*a. True  
b. False  
94. Algae is primarily responsible for the effective working of an anaerobic pond  
a. True  
*b. False  
95. Tertiary (polishing) ponds typically receive raw (untreated wastewater) as the feed stream  
a. True  
*b. False  
96. In a facultative pond, a drop in the pH will normally be accompanied by a rise in the dissolved  
oxygen.  
*a. True  
10.4 Ponds Operations and Maintenance  
151  
b. False  
97. One short-term corrective measure for an overloaded facultative pond might be to add:  
a. copper sulfate .  
b. sodium sulfide.  
c. ammonium sulfide .  
*d. sodium nitrate.  
e. potassium chloride.  
98. Photosynthesis is an essential part of the biological activity associated with:  
a. Activated sludge  
b. Trickling filters  
c. Oxidation ditches  
d. Aerobic digesters  
*e. Sewage lagoons  
99. During the process of algal photosynthesis:  
a. Chlorophyll converts sunlight into energy for growth.  
b. Algae produces oxygen  
c. Algae converts CO2, NH3, and PO4, into additional algae cells  
*d. All of the above  
100. pH of the facultative pond will be the highest  
*a. during daytime when the consumption of CO2 is the highest  
b. during daytime when the consumption of CO2 is the lowest  
c. during nighttime when the production of CO2 is highest  
d. during nighttime when the production of CO2 is lowest  
101. A lagoon operator collects a sample of effluent at 2:15 pm. on a sunny July day and tests it. for  
dissolved oxygen. The dissolved oxygen is 22 mg/1 and the pH is 9.2. The lagoon has a green  
color. Effluent suspended solids have been running at 75 mg/1. The operator should  
a. Do nothing. The conditions described are normal.  
*b. Apply algaecide to the lagoon to kill the algae.  
c. Drawdown the lagoon to eliminate excess DO.  
d. Isolate cell.  
102. Algae in a stabilization pond is most likely to:  
a. consume oxygen during daylight hours.  
b. decrease effluent TSS during the day.  
*c. change the pH throughout the day.  
d. increase oxygen at night.  
e. None of the above.  
103. Algae in a stabilization pond is most likely to:  
a. consume oxygen during daylight hours.  
b. decrease effluent TSS during the day.  
*c. change the pH throughout the day.  
d. increase oxygen at night.  
e. None of the above.  
104. An operator cannot maintain adequate water levels in one of the ponds. What will cause this to  
happen?  
a. The pond is hydraulically overloaded.  
152  
Chapter 10. Stabilzation Ponds  
*b. The pond seal leaks.  
c. The flow control structure leaks.  
d. The flow control structure does not split the flow evenly.  
105. A properly designed and operated wastewater stabilization pond will remove  
percent  
BOD.  
a. 40%-60%.  
b. 60%-70%.  
c. 70%-80%.  
*d. 80%-90%.  
106. At night algae in a conventional lagoon will:  
*a. Cease to produce oxygen  
b. Consume oxygen  
c. Produce less oxygen  
d. Increase the pH of the lagoon contents  
107. At what time of day is the dissolved oxygen content highest in a lagoon?  
a. 3 a.m.  
b. 7 a.m.  
c. 9 a.m.  
*d. 3 p.m.  
108. Cattails growing in lagoon will  
*a. Cause short circuiting in affected lagoon.  
b. Eliminate mosquito larvae.  
c. Increase the diurnal pH fluctuations.  
d. Increase toxic blue-green algae concentrations in the effluent.  
109. Dike vegetation should be controlled by  
*a. Mowing periodically.  
b. Burning in the spring and fall.  
c. Allowing the cattle to graze on the dikes.  
d. Any of the above would be acceptable.  
110. Due to diurnal differences in operation, a lagoon system is likely to experience the lowest dissolved  
oxygen readings  
a. At any time since diurnal differences have no bearing on DO values  
b. During the time when the sun is out and it is the hottest  
*c. During night, just before dawn  
d. When BOD loading is the lowest  
111. During the process of algal photosynthesis:  
a. Chlorophyll converts sunlight into energy for growth.  
b. Algae produces oxygen  
c. Algae converts CO2, NH3, and PO4, into additional algae cells  
*d. All of the above  
112. Given the following data, what is the most likely cause of the aerated pond problem? — DATA:  
Turbulence over center part of pond. Solids floating to surface on edge of pond. DO in center  
portion= 1.4 mg/L.  
a. Hydraulic flow too slow through pond.  
*b. Inadequate aeration on edges on pond.  
10.4 Ponds Operations and Maintenance  
153  
c. Sludge scraper not operating.  
d. Toxic materials in pond.  
113. Hydraulic loading to a facultative pond equals:  
a. Volume (MG) divided by Flow (MGD)  
*b. Depth (inches) divided by Detention Time (days)  
c. Volume (acre-feet) divided by Flow (acre-inches/day)  
d. Flow ( gallons/day) divided by Area (acres)  
e. Area (acres) divided by Flow (acre-inches/day)  
114. In a properly operating facultative pond, algae live on carbon dioxide and nutrients during the day,  
and at night produce carbon dioxide. This has what effect on the pH?  
*a. pH increases during the day, and decreases at night  
b. pH decreases during the day, and increases at night  
c. pH stays the same no matter what time of day  
d. Carbon dioxide has no effect on pH  
115. It is important to completely mix and aerate the following type of pond:  
*a. aerobic.  
b. facultative.  
c. anaerobic.  
d. All of the above.  
e. None of the above.  
116. One short-term corrective measure for an overloaded facultative pond might be to add:  
a. copper sulfate .  
b. sodium sulfide.  
c. ammonium sulfide .  
*d. sodium nitrate.  
e. potassium chloride.  
117. pH of the facultative pond will be the highest  
*a. during daytime when the consumption of CO2 is the highest  
b. during daytime when the consumption of CO2 is the lowest  
c. during nighttime when the production of CO2 is highest  
d. during nighttime when the production of CO2 is lowest  
118. Photosynthesis is an essential part of the biological activity associated with:  
a. Activated sludge  
b. Trickling filters  
c. Oxidation ditches  
d. Aerobic digesters  
*e. Sewage lagoons  
119. Due to diurnal differences in operation, a facultative pond is likely to experience the lowest dis-  
solved oxygen readings  
a. At any time since diurnal differences have no bearing on DO values  
b. During the time when the sun is out and it is the hottest  
*c. During night, just before dawn  
d. When BOD loading is the lowest  
120. Due to diurnal differences in operation, a facultative pond is likely to experience the lowest dis-  
solved oxygen readings  
154  
Chapter 10. Stabilzation Ponds  
a. At any time since diurnal differences have no bearing on DO values  
b. During the time when the sun is out and it is the hottest  
*c. During night, just before dawn  
d. When BOD loading is the lowest  
121. The main function of algae in a facultative wastewater treatment pond is to:  
a. produce carbon dioxide, which is then used by the facultative bacteria.  
b. use up nutrients such as nitrogen and phosphorus.  
*c. produce oxygen during daylight hours.  
d. serve as a food source for essential protozoa  
e. break down complex organics in the wastewater.  
122. Which of the following microorganisms are involved in the stabilization of wastewater in a faculta-  
tive wastewater treatment pond?  
a. Aerobic bacteria  
b. Anaerobic  
c. Facultative bacteria  
*d. All of the above  
e. (a) and (c) only  
123. Which of the following terms is usually not associated with wastewater treatment ponds?  
*a. Filamentous bacteria  
b. Symbiotic relationship  
c. Photosynthesis  
d. Sewage lagoon  
e. Inches/day  
124. Which of the following statements is not true regarding a facultative wastewater treatment pond?  
a. When starting a facultative pond, 1 foot of relatively clean water should be added to the pond  
prior to the addition of wastewater.  
b. A facultative pond generally is operated at a detention time of 50 to 60 days or longer.  
*c. DO concentrations of 10 to 15 mg/L or greater are frequently found in a facultative pond during  
the afternoon of a sunny day.  
d. Organic loading to a pond is expressed as pounds of volatile suspended solids per acre per day.  
e. A facultative pond has an anaerobic layer and an aerobic layer.  
125. The main function of algae in a facultative wastewater treatment pond is to:  
a. produce carbon dioxide, which is then used by the facultative bacteria.  
b. use up nutrients such as nitrogen and phosphorus.  
*c. produce oxygen during daylight hours.  
d. serve as a food source for essential protozoa  
e. break down complex organics in the wastewater.  
126. Which of the following terms is usually not associated with wastewater treatment ponds?  
*a. Filamentous bacteria  
b. Symbiotic relationship  
c. Photosynthesis  
d. Sewage lagoon  
e. Inches/day  
127. Which of the following statements is not true regarding a facultative wastewater treatment pond?  
a. When starting a facultative pond, 1 foot of relatively clean water should be added to the pond  
10.4 Ponds Operations and Maintenance  
155  
prior to the addition of wastewater.  
b. A facultative pond generally is operated at a detention time of 50 to 60 days or longer.  
*c. DO concentrations of 10 to 15 mg/L or greater are frequently found in a facultative pond during  
the afternoon of a sunny day.  
d. Organic loading to a pond is expressed as pounds of volatile suspended solids per acre per day.  
e. A facultative pond has an anaerobic layer and an aerobic layer.  
128. Which of the following would not be a routine operational or maintenance problem in the operation  
of pond:  
a. weed control.  
b. levee maintenance.  
c. insect control.  
*d. temperature control.  
e. scum control.  
129. Algae in a stabilization pond is most likely to:  
a. consume oxygen during daylight hours.  
b. decrease effluent TSS during the day.  
*c. change the pH throughout the day.  
d. increase oxygen at night.  
e. None of the above.  
130. At night algae in a conventional lagoon will:  
*a. Cease to produce oxygen  
b. Consume oxygen  
c. Produce less oxygen  
d. Increase the pH of the lagoon contents  
131. Which of the following microorganisms are involved in the stabilization of wastewater in a faculta-  
tive wastewater treatment pond?  
a. Aerobic bacteria  
b. Anaerobic  
c. Facultative bacteria  
*d. All of the above  
e. (a) and (c) only  
132. Which of the following is not a term used to refer to a conventional wastewater treatment lagoon?  
a. Oxidation pond  
b. Stabilization pond  
c. Facultative lagoon  
*d. Aerobic lagoon  
133. The main function of algae in a facultative wastewater treatment pond is to:  
a. produce carbon dioxide, which is then used by the facultative bacteria.  
b. use up nutrients such as nitrogen and phosphorus.  
*c. produce oxygen during daylight hours.  
d. serve as a food source for essential protozoa.  
e. break down complex organics in the wastewater.  
134. Sodium nitrate is a chemical that may be used in the operation of a wastewater pond to restore  
normal conditions after a pond "turn over." Its function is:  
*a. To provide a source of chemically combined oxygen for the facultative bacteria.  
156  
Chapter 10. Stabilzation Ponds  
b. To kill off the unwanted blue-green algae.  
c. To preserve the nitrogen balance in the pond.  
d. To control the growth of weeds.  
e. To neutralize chlorine residuals in the pond effluent.  
135. It is important to completely mix and aerate the following type of pond:  
*a. aerobic.  
b. facultative.  
c. anaerobic.  
d. All of the above.  
e. None of the above.  
136. The laboratory tests most frequently used to monitor a facultative pond on a daily basis are:  
a. pH, color, and BOD  
*b. pH, BOD, and suspended solids  
c. pH, DO, and temperature  
d. DO, BOD, and SS  
e. Microscopic examination, color, and DO  
137. The effluent from a conventional lagoon should be withdrawn:  
a. Off the surface  
b. Near the bottom  
*c. Six to eighteen inches below the surface  
d. Intermittently  
138. Algae in a stabilization pond is most likely to:  
a. consume oxygen during daylight hours.  
b. decrease effluent TSS during the day.  
*c. change the pH throughout the day.  
d. increase oxygen at night.  
e. None of the above.  
139. Which of the following terms is usually not associated with wastewater treatment ponds?  
*a. Filamentous bacteria  
b. Symbiotic relationship  
c. Photosynthesis  
d. Sewage lagoon  
e. Inches/day  
140. Which of the following statements is not true regarding a facultative wastewater treatment pond?  
a. When starting a facultative pond, 1 foot of relatively clean water should be added to the pond  
prior to the addition of wastewater.  
b. A facultative pond generally is operated at a detention time of 50 to 60 days or longer.  
*c. DO concentrations of 10 to 15 mg/L or greater are frequently found in a facultative pond during  
the afternoon of a sunny day.  
d. Organic loading to a pond is expressed as pounds of volatile suspended solids per acre per day.  
e. A facultative pond has an anaerobic layer and an aerobic layer.  
141. One short-term corrective measure for an overloaded facultative pond might be to add:  
a. copper sulfate .  
b. sodium sulfide.  
c. ammonium sulfide .  
10.4 Ponds Operations and Maintenance  
157  
*d. sodium nitrate.  
e. potassium chloride.  
142. Photosynthesis is an essential part of the biological activity associated with:  
a. Activated sludge  
b. Trickling filters  
c. Oxidation ditches  
d. Aerobic digesters  
*e. Sewage lagoons  
143. pH of the facultative pond will be the highest  
*a. during daytime when the consumption of CO2 is the highest  
b. during daytime when the consumption of CO2 is the lowest  
c. during nighttime when the production of CO2 is highest  
d. during nighttime when the production of CO2 is lowest  
144. One short-term corrective measure for an overloaded facultative pond might be to add:  
a. copper sulfate .  
b. sodium sulfide.  
c. ammonium sulfide .  
*d. sodium nitrate.  
e. potassium chloride.  
145. Photosynthesis is an essential part of the biological activity associated with:  
a. Activated sludge  
b. Trickling filters  
c. Oxidation ditches  
d. Aerobic digesters  
*e. Sewage lagoons  
146. During the process of algal photosynthesis:  
a. Chlorophyll converts sunlight into energy for growth.  
b. Algae produces oxygen  
c. Algae converts CO2, NH3, and PO4, into additional algae cells  
*d. All of the above  
147. pH of the facultative pond will be the highest  
*a. during daytime when the consumption of CO2 is the highest  
b. during daytime when the consumption of CO2 is the lowest  
c. during nighttime when the production of CO2 is highest  
d. during nighttime when the production of CO2 is lowest  
148. A lagoon operator collects a sample of effluent at 2:15 pm. on a sunny July day and tests it. for  
dissolved oxygen. The dissolved oxygen is 22 mg/1 and the pH is 9.2. The lagoon has a green  
color. Effluent suspended solids have been running at 75 mg/1. The operator should  
a. Do nothing. The conditions described are normal.  
*b. Apply algaecide to the lagoon to kill the algae.  
c. Drawdown the lagoon to eliminate excess DO.  
d. Isolate cell.  
149. Algae in a stabilization pond is most likely to:  
a. consume oxygen during daylight hours.  
b. decrease effluent TSS during the day.  
158  
Chapter 10. Stabilzation Ponds  
*c. change the pH throughout the day.  
d. increase oxygen at night.  
e. None of the above.  
150. An operator cannot maintain adequate water levels in one of the ponds. What will cause this to  
happen?  
a. The pond is hydraulically overloaded.  
*b. The pond seal leaks.  
c. The flow control structure leaks.  
d. The flow control structure does not split the flow evenly.  
151. A properly designed and operated wastewater stabilization pond will remove what percent BOD.  
a. 40%-60%.  
b. 60%-70%.  
c. 70%-80%.  
*d. 80%-90%.  
152. At night algae in a conventional lagoon will:  
*a. Cease to produce oxygen  
b. Consume oxygen  
c. Produce less oxygen  
d. Increase the pH of the lagoon contents  
153. At what time of day is the dissolved oxygen content highest in a lagoon?  
a. 3 a.m.  
b. 7 a.m.  
c. 9 a.m.  
*d. 3 p.m.  
154. Cattails growing in lagoon will  
*a. Cause short circuiting in affected lagoon.  
b. Eliminate mosquito larvae.  
c. Increase the diurnal pH fluctuations.  
d. Increase toxic blue-green algae concentrations in the effluent.  
155. Dike vegetation should be controlled by  
*a. Mowing periodically.  
b. Burning in the spring and fall.  
c. Allowing the cattle to graze on the dikes.  
d. Any of the above would be acceptable.  
156. Due to diurnal differences in operation, a lagoon system is likely to experience the lowest dissolved  
oxygen readings  
a. At any time since diurnal differences have no bearing on DO values  
b. During the time when the sun is out and it is the hottest  
*c. During night, just before dawn  
d. When BOD loading is the lowest  
157. Given the following data, what is the most likely cause of the aerated pond problem? —  
DATA: Turbulence over center part of pond. Solids floating to surface on edge of pond. DO in  
center portion= 1.4 mg/L.  
a. Hydraulic flow too slow through pond.  
*b. Inadequate aeration on edges on pond.  
10.4 Ponds Operations and Maintenance  
159  
c. Sludge scraper not operating.  
d. Toxic materials in pond.  
158. Hydraulic loading to a facultative pond equals:  
a. Volume (MG) divided by Flow (MGD)  
*b. Depth (inches) divided by Detention Time (days)  
c. Volume (acre-feet) divided by Flow (acre-inches/day)  
d. Flow ( gallons/day) divided by Area (acres)  
e. Area (acres) divided by Flow (acre-inches/day)  
159. In a properly operating facultative pond, algae live on carbon dioxide and nutrients during the day,  
and at night produce carbon dioxide. This has what effect on the pH?  
*a. pH increases during the day, and decreases at night  
b. pH decreases during the day, and increases at night  
c. pH stays the same no matter what time of day  
d. Carbon dioxide has no effect on pH  
160. It is important to completely mix and aerate the following type of pond:  
*a. aerobic.  
b. facultative.  
c. anaerobic.  
d. All of the above.  
e. None of the above.  
161. Due to diurnal differences in operation, a facultative pond is likely to experience the lowest dis-  
solved oxygen readings  
a. At any time since diurnal differences have no bearing on DO values  
b. During the time when the sun is out and it is the hottest  
*c. During night, just before dawn  
d. When BOD loading is the lowest  
162. The main function of algae in a facultative wastewater treatment pond is to:  
a. produce carbon dioxide, which is then used by the facultative bacteria.  
b. use up nutrients such as nitrogen and phosphorus.  
*c. produce oxygen during daylight hours.  
d. serve as a food source for essential protozoa  
e. break down complex organics in the wastewater.  
163. Which of the following microorganisms are involved in the stabilization of wastewater in a faculta-  
tive wastewater treatment pond?  
a. Aerobic bacteria  
b. Anaerobic  
c. Facultative bacteria  
*d. All of the above  
e. (a) and (c) only  
164. Which of the following statements is not true regarding a facultative wastewater treatment pond?  
a. When starting a facultative pond, 1 foot of relatively clean water should be added to the pond  
prior to the addition of wastewater.  
b. A facultative pond generally is operated at a detention time of 50 to 60 days or longer.  
*c. DO concentrations of 10 to 15 mg/L or greater are frequently found in a facultative pond during  
the afternoon of a sunny day.  
160  
Chapter 10. Stabilzation Ponds  
d. Organic loading to a pond is expressed as pounds of volatile suspended solids per acre per day.  
e. A facultative pond has an anaerobic layer and an aerobic layer.  
165. Which of the following would not be a routine operational or maintenance problem in the operation  
of pond:  
a. weed control.  
b. levee maintenance.  
c. insect control.  
*d. temperature control.  
e. scum control.  
166. Which of the following is not a term used to refer to a conventional wastewater treatment lagoon?  
a. Oxidation pond  
b. Stabilization pond  
c. Facultative lagoon  
*d. Aerobic lagoon  
167. Sodium nitrate is a chemical that may be used in the operation of a wastewater pond to restore  
normal conditions after a pond "turn over." Its function is:  
*a. To provide a source of chemically combined oxygen for the facultative bacteria.  
b. To kill off the unwanted blue-green algae.  
c. To preserve the nitrogen balance in the pond.  
d. To control the growth of weeds.  
e. To neutralize chlorine residuals in the pond effluent.  
168. The laboratory tests most frequently used to monitor a facultative pond on a daily basis are:  
a. pH, color, and BOD  
*b. pH, BOD, and suspended solids  
c. pH, DO, and temperature  
d. DO, BOD, and SS  
e. Microscopic examination, color, and DO  
169. The effluent from a conventional lagoon should be withdrawn:  
a. Off the surface  
b. Near the bottom  
*c. Six to eighteen inches below the surface  
d. Intermittently  
170. The facultative zone in a pond:  
a. provides no treatment.  
b. provides oxygen to the anaerobic zone.  
c. is a barrier to light penetration  
d. produces the bacteria for the anaerobic zone.  
*e. uses nitrate as an oxygen source for digestion.  
171. Which of the following terms is usually not associated with wastewater treatment ponds?  
*a. Filamentous bacteria  
b. Symbiotic relationship  
c. Photosynthesis  
d. Sewage lagoon  
e. Inches/day  
172. A lagoon operator collects a sample of effluent at 2:15 pm. on a sunny July day and tests it. for  
10.4 Ponds Operations and Maintenance  
161  
dissolved oxygen. The dissolved oxygen is 22 mg/1 and the pH is 9.2. The lagoon has a green  
color. Effluent suspended solids have been running at 75 mg/1. The operator should  
a. Do nothing. The conditions described are normal.  
*b. Apply algaecide to the lagoon to kill the algae.  
c. Drawdown the lagoon to eliminate excess DO.  
d. Isolate cell.  
173. Given the following data, what is the most likely cause of the aerated pond problem? —  
DATA: Turbulence over center part of pond. Solids floating to surface on edge of pond. DO in  
center portion= 1.4 mg/L.  
a. Hydraulic flow too slow through pond.  
*b. Inadequate aeration on edges on pond.  
c. Sludge scraper not operating.  
d. Toxic materials in pond.  
174. Hydraulic loading to a facultative pond equals:  
a. Volume (MG) divided by Flow (MGD)  
*b. Depth (inches) divided by Detention Time (days)  
c. Volume (acre-feet) divided by Flow (acre-inches/day)  
d. Flow ( gallons/day) divided by Area (acres)  
e. Area (acres) divided by Flow (acre-inches/day)  
175. Which of the following is not a term used to refer to a conventional wastewater treatment lagoon?  
a. Oxidation pond  
b. Stabilization pond  
c. Facultative lagoon  
*d. Aerobic lagoon  
176. Sodium nitrate is a chemical that may be used in the operation of a wastewater pond to restore  
normal conditions after a pond "turn over." Its function is:  
*a. To provide a source of chemically combined oxygen for the facultative bacteria.  
b. To kill off the unwanted blue-green algae.  
c. To preserve the nitrogen balance in the pond.  
d. To control the growth of weeds.  
e. To neutralize chlorine residuals in the pond effluent.  
177. The laboratory tests most frequently used to monitor a facultative pond on a daily basis are:  
a. pH, color, and BOD  
*b. pH, BOD, and suspended solids  
c. pH, DO, and temperature  
d. DO, BOD, and SS  
e. Microscopic examination, color, and DO  
178. The effluent from a conventional lagoon should be withdrawn:  
a. Off the surface  
b. Near the bottom  
*c. Six to eighteen inches below the surface  
d. Intermittently  
179. Sodium nitrate is a chemical that may be used in the operation of a wastewater pond to restore  
normal conditions after a pond "turn over." Its function is:  
*a. To provide a source of chemically combined oxygen for the facultative bacteria.  
162  
Chapter 10. Stabilzation Ponds  
b. To kill off the unwanted blue-green algae.  
c. To preserve the nitrogen balance in the pond.  
d. To control the growth of weeds.  
e. To neutralize chlorine residuals in the pond effluent.  
180. The effluent from a conventional lagoon should be withdrawn:  
a. Off the surface  
b. Near the bottom  
*c. Six to eighteen inches below the surface  
d. Intermittently  
181. The facultative zone in a pond:  
a. provides no treatment.  
b. provides oxygen to the anaerobic zone.  
c. is a barrier to light penetration.  
d. produces the bacteria for the anaerobic zone.  
*e. uses nitrate as an oxygen source for digestion.  
182. One short-term corrective measure for an overloaded facultative pond might be to add:  
a. copper sulfate .  
b. sodium sulfide.  
c. ammonium sulfide .  
*d. sodium nitrate.  
e. potassium chloride.  
183. During the process of algal photosynthesis:  
a. Chlorophyll converts sunlight into energy for growth.  
b. Algae produces oxygen  
c. Algae converts CO2, NH3, and PO4, into additional algae cells  
*d. All of the above  
184. pH of the facultative pond will be the highest  
*a. during daytime when the consumption of CO2 is the highest  
b. during daytime when the consumption of CO2 is the lowest  
c. during nighttime when the production of CO2 is highest  
d. during nighttime when the production of CO2 is lowest  
11. Activated Sludge  
11.1 Process overview  
Activated sludge is a secondary, biological treatment process used for the removal of suspended and  
dissolved organic matter from the primary treated wastewater.  
• Utilizes an aeration basin/reactor and a secondary clarifier  
In the presence of oxygen, aerobic bacteria in the aeration basin consume the organic matter (BOD)  
in wastewater for their growth and reproduction, converting BOD into bacterial cell mass along  
with metabolic byproducts including carbon dioxide and water  
The aerobic bacteria is the predominant microbial life form in the aeration basin. Other higher  
microbial life forms — mainly protozoa, are present along with some metazoans.  
The microorganisms along with their metabolic byproducts and residual dead cell mass form a  
cluster called floc.  
The wastewater exiting the aeration basin enters a clarifier where the floc settles. The clear, treated  
secondary effluent flows out.  
A portion of the settled activated sludge floc, is returned from the clarifier to the front of the aera-  
tion basin to seed the activated sludge treatment of the incoming primary effluent. The recycled floc  
is called Return Activated Sludge (RAS).  
The remaining settled floc from the clarifier is ”wasted” —transferred for solids treatment (typ-  
ically using digestion) prior to its ultimate disposal. The wasted floc is called Waste Activated  
Sludge (WAS).  
• The color of healthy activated sludge is tan to brown with an earthy/musty odor.  
 
 
164  
Chapter 11. Activated Sludge  
For activated sludge treatment to be effective, it is critical to establish a healthy microbial popula-  
tion which converts the BOD into easily separable biomass.  
If the biomass does not settle well in the clarifier, it will be carried out in the treated secondary  
effluent producing a poor quality effluent with higher solids and organic content.  
Mixed liquor:  
• Mixed liquor is the mixture of raw and/or treated wastewater with the biomass.  
• Mixed liquor sample is typically taken at the end of the aeration reactor.  
In the mixed liquor, the suspended solids comprised mostly of biomass with some undegraded and  
non-degradable material is called Mixed liquor suspended solids (MLSS).  
• The volatile fraction of the MLSS is termed Mixed Liquor Volatile Solids (MLVSS).  
The MLVSS is the amount of biomass as a percentage of the MLSS. Generally, the MLVSS is 70  
- 80%, implying that 70-80% of the MLSS is volatile (organic) solids.  
11.2 Process parameters  
11.2.1 Dissolved oxygen (DO)  
Maintaining adequate Dissolved oxygen (DO) is key to the activated sludge process. DO is typically  
maintained between 0.5 to 3.0 mg/l in the aeration reactor.  
11.2.2 pH  
The microbiological population in the activated sludge is very sensitive to pH and needs to be maintained  
at neutral - 7 pH.  
11.2.3 Temperature  
Biological activity is temperature dependent. Microbial activity approximately doubles for every 10oC  
rise in temperature.  
11.2.4 Oxygen uptake rate (OUR) and specific oxygen uptake rate (SOUR)  
Oxygen Uptake Rate (OUR) involves measurement of the amount of oxygen used up by the microor-  
ganisms using a DO probe and is expressed in unit time of mg/L-hr (ppm O2 consumed per hour). By  
knowing the OUR, we can establish the activity of the microorganisms in the aeration tank and know if  
they are consuming the oxygen provided for removing organic matter. For conventional activated sludge  
process the typical OUR values range from 10 to 30 mg/L-hr  
Specific Oxygen Uptake Rate (SOUR) provides the OUR information based on the concentration of  
microorganisms present. SOUR is obtained by dividing OUR with MLVSS. The value is indicated and  
 
 
 
 
 
11.2 Process parameters  
mg  
165  
O2  
mg  
gm  
l hr  
measured in terms of unit of  
1000  
.
mg  
l
MLVSS  
Optimal range of SOUR is usually between 8 to 20.  
If a baseline OUR is established for a system - presence of toxic susbstances in that system can be identi-  
fied by a drop in OUR. SOUR value above the optimal range indicates an increasing F:M - Higher BOD  
load with less than adequate microorganisms (represented by MLVSS) indicating a young sludge and the  
potential for straggler floc. A lower than optimal SOUR value indicates insufficient food to support the  
microorganisms indicating an older sludge and the potential for pin floc.  
11.2.5 Food to Mass ratio (F:M)  
In order to establish the amount of the MLSS to be maintained in the aeration tank, the parameter Food  
to Mass Ratio (F:M) is used. F:M is the ratio of the "food" – the mass of primary effluent BOD entering  
the aeration basin to the mass of the microorganisms in the aeration basin —MLVSS. Common ranges for  
F:M for a conventional activated sludge plant are from 0.15 to 0.5  
11.2.6 Sludge age  
• Sludge age is the average time (days) that a sludge particle (cell) spends in the system.  
Sludge age dictates the presence and predominance of the different types of organisms and is  
measured as Mean Cell Retention Time (MCRT) or Solids Retention Time (SRT).  
• Sludge age is the average time that a sludge particle (cell) spends in the system  
• Conventional AS process - MCRTs are in the range of 5 to 15 days  
• The desired MCRT for a given plant must be found experimentally  
F:M and MCRT are inversely related: that is a long MCRT means a low F:M and a short MCRT  
means a high F:M  
Both, F:M and MCRT values vary from plant to plant and are established through a trial and error  
process.  
• The MCRT is calculated as:  
MCRT(days) =  
Total MLSS lbs in the aeration system (aeration tank  
+ clarifier)  
Total amount in lbs/day of suspended solids leaving the system (E f fluent SS+WAS solids)  
MLSS in aeration tank (lbs)+MLSS in clarifier (lbs)  
MCRT(days) =  
E f fluent suspended solids (lbs/day)+ WAS SS (lbs/day)  
11.2.7 Sludge settleability  
A settleability test is conducted on the mixed liquor from the aeration reactor to measure the com-  
paction and settleability characteristics of the secondary solids.  
In the settleability test, a 1-liter mixed liquor sample is allowed to settle for 30-minutes in a gradu-  
ated cylinder settleometer.  
The Sludge Volume Index (SVI) of the sludge is calculated using results from the 30-minute  
settleability test and the MLSS value.  
SVI is expressed in ml/g and it is essentially the volume (ml) of 1 gram of the MLSS after 30  
minutes of settling.  
Settled sludge volume in ml/l after 30 min  
• SVI (ml/g)=  
1000mg  
g
MLSS mg/l  
SVI provides a more accurate picture of the sludge settling characteristics than settleability or  
 
 
 
166  
Chapter 11. Activated Sludge  
MLSS alone.  
• 50 to 120 ml/g SVI value is considered optimal.  
• Higher SVI values indicate sludge that is slow to settle and not compacting well.  
• When SVI values are approaching 200 ml/g, activated sludge process is considered to be "bulking".  
Regular monitoring of SVI help identify changes occurring in the activated sludge process prevent-  
ing settling problems before they occur.  
Like F:M and MCRT, the optimal SVI value for each plant varies and is also established by trial  
and error.  
11.3 Microbiology  
Bacteria are fundamental microorganisms in the stabilization of organic wastes and therefore of basic  
importance in the secondary wastewater treatment process.  
Bacteria (singular, bacterium) are single celled organisms and represent the simplest forms of life. These  
organisms utilize soluble food and are capable of self-reproduction. Bacteria can be classified in many  
different ways.  
For wastewater bacteria are classified based on nutritive requirement, bacteria are classified as het-  
erotrophic or autotrophic bacteria, although several species may function both.  
1. Heterotrophic bacteria use carbon based organic compounds for both energy and food source. A  
term commonly used instead of heterotroph is “saprophyte” which refers to an organism that lives  
on dead or decaying organic matter. The heterotrophic bacteria are grouped into three classifica-  
tions, depending upon their action toward free oxygen.  
a. Aerobes : Require free dissolved oxygen to live and multiply.  
b. Anaerobes : Oxidize organic matter in the complete absence of dissolved oxygen.  
c. Facultative : Use free dissolved oxygen when available but can also respire and multiply in  
its absence, e.g. Escheichia coli.  
2. Autotrophic bacteria use carbon dioxide as a carbon source and oxidize inorganic compounds for  
energy. Autotrophs of greatest significance in wastewater treatment are the nitrifying and sulfur  
bacteria.  
a. Nitrifying bacteria : These bacteria are responsible for the oxidation of ammonia in wastewa-  
ter to nitrates.  
b. Sulfur bacteria : These bacteria are involved in the formation of sulfuric acid from the dis-  
solved sulfides in the sewer conveyance systems.  
The activated sludge is a complex ecosystem of aerobic microorganisms predominantly heterotrophic  
bacteria along with some autotrophic bacteria and other trophic organisms including protozoas, rotifers and  
nematodes which feed on the bacteria and particulate matter.  
Protozoa are single-celled microscopic organisms, several hundred times larger than bacteria. It is the  
protozoa we observe under a microscope since bacteria are actually too small to see. There are four types of  
protozoa commonly found in activated sludge systems. They are identified by their method of movement  
within the wastewater environment. The four types are:  
 
11.3 Microbiology  
167  
Amoeba  
StalkedCilliate  
Flagellate  
Suctorian  
Other commonly found organisms included the multi-celled organisms - Metazoans such as Rotifers and  
Nematodes  
Rotifer  
Nematode  
The essence of the activated sludge system is to develop and and maintain a diverse population of  
microorganisms (activated sludge) that treats wastewater and which can be managed.  
The activated sludge should be such that it not only effectively remove the BOD, but also produces a  
168  
Chapter 11. Activated Sludge  
floc that settles and compacts well in the secondary clarifier thereby producing a secondary effluent  
with a low BOD content.  
The floc produced has a porous structure and is essentially an aggregate of bacteria and extracelu-  
lar material including organic polymers - polysaccharides and other substances secreted by the  
microorganisms, and remnants of dead microorganisms.  
Typically, the trophic organisms such as the protozoas and rotifers which feed on the bacteria and  
other particulates are generally present on the outer surface of the flocs.  
• A good quality activated sludge is brown in color with some froth and a slight musty odor.  
The settleability of the sludge is measured as SVI which is the volume of settled sludge in milliliters  
occupied by 1 gram of dry sludge solids after 30 minutes of settling in a 1,000 mL graduated cylin-  
der or a settleometer.  
• SVI of a good settling sludge is typically between 50 - 150 ml/g.  
• MCRT is the key factor which establishes the organism mix in the activated sludge.  
• As MCRT increases, the size and complexity of the organisms increases.  
The types of organisms predominating in the mixed liquor give an indication as to the age of the  
sludge.  
1. At low MCRT of about less than 4 days the simpler protozoas - amoebae and flagellates  
dominate.  
2. With an increase in MCRT more complex organisms such as the free swimming ciliates and  
stalked ciliates appear.  
3. At even higher MCRT, metazoans such as the rotifers and nematodes may be found  
11.4 Activated sludge process control  
To ensure optimal treatment in the activated sludge process - maximum BOD removal with a well  
settling mixed liquor, controlling the inventory of solids (biomass) is a key process control parame-  
ter.  
New solids (biomass) are produced in the activated sludge reactors due to the consumption of the  
BOD by the microorganisms.  
• The solids faction of the mixed liquor which separate (settle) in the secondary clarifier are  
Either returned to the influent end of the activated sludge reactor as Return Activated Sludge  
(RAS), or  
removed from the system as Waste Activated Sludge (WAS).  
Thus, RAS and WAS pumping controls are used for maintaining the desired amount of solids in the  
system.  
11.4.1 Controlling waste activated sludge (WAS) pumping rate  
For process stability and control the wasting should be continuous and not intermittent and any increase  
or decrease in wasting should be made gradually, i.e., 20 - 25 percent per day. One of the following ap-  
proaches is utilized to establish the WAS flow control  
Constant Mean Cell Residence Time (MCRT)  
Typical MCRT for conventional AS process ranges from 5 to 15 days and 20 to 30 days for extended  
aeration and for a given plant, the desired MCRT is established based upon operational experience. WAS  
flow control can be established to maintain the desired MCRT range for a given plant.  
 
 
11.4 Activated sludge process control  
169  
Constant food:mass (F/M)  
This parameter is based upon the ratio of food (influent BOD) fed to the microorganisms each day to the  
mass of microorganisms held under aeration. WAS flow control can be used for controlling the mass of  
microorganisms in the system to maintain the desired F/M ratio.  
Constant Mixed Liquor Suspended Solids (MLSS)  
A MLSS concentration which provides the best effluent and highest removal efficiencies is selected and  
WAS flow control is established to maintain the desired MLSS concentration level. Wasting is increased  
when the MLSS concentration is higher and decreased or stopped when the MLSS concentration is lower  
than the desired level.  
11.4.2 Controlling the return activated sludge (RAS) rate  
RAS flow rate affects the concentration and the sludge age. Return Activated Sludge Control... To prop-  
erly operate the activated sludge process, For conventional activated sludge operations, the RAS flow is  
generally about 30 to 100% of the incoming wastewater flow. Changes in the activated sludge quality will  
require different RAS flow rates due to settling characteristics of the sludge. The following approaches are  
used to control the RAS flow rate:  
Constant RAS flow rate control - RAS pumping independent of influent flow  
Results in a continuously varying MLSS concentration that will be at a minimum during peak  
influent flows and a maximum during minimum influent flows. This occurs because the MLSS are  
flowing into the clarifier at a lower rate during peak flow when being removed at a constant rate.  
Similarly, at minimum influent flow rates, the MLSS are being returned to the aeration tank at a  
higher rate than are flowing into the clarifier  
• Maximum solids loading on the clarifier occurs at the initial start of peak flow periods.  
The clarifier will have has a constantly changing depth of sludge blanket as the MLSS moves from  
the aeration tank to the clarifier and vice versa.  
This RAS control option is especially advantageous for smaller plants as it entails little effort/minimal  
automation for RAS pumping control.  
A disadvantage of using the constant flow approach is that the F/M is constantly changing. The  
range of F/M fluctuation due to the effect of short term variation in the MLSS (because of hydraulic  
loading) is generally small enough so that no significant problems arise due to using this approach.  
Constant percentage RAS flow rate control  
This approach requires a programmed method for maintaining a constant percentage of the aeration  
tank influent wastewater flow rate. The program may consist of an automatic flow measurement  
device, a programmed system, or frequent manual adjustments.  
The variations in the MLSS concentration due to the diurnal flow profile are reduced and the F/M  
varies less.  
The MLSS will remain in the clarifier for shorter time periods, which may reduce the possibility of  
denitrification in the clarifier.  
Clarifier is subjected to maximum solids loading when the clarifier contains the maximum amount  
of sludge. This may result in solids washout with the effluent.  
 
170  
Chapter 11. Activated Sludge  
Establishing RAS flow rate based on a solids balance approach  
The following equation can be used for controlling RAS flow rate to maintain a target MLSS concentra-  
tion given the influent flow and the RAS solids content.  
Applying a mass balance approach to the secondary clarifier:  
Mass of solids in = Mass of solids Out  
(Q+QRAS)MLSS = QRAS SSRAS +QWAS SSRAS  
Notes:  
• The solids leaving with the effluent - QSSEFF is negligible  
• Also, SSWAS = SSRAS  
Rearranging:  
 
!
QMLSSQWAS SSRAS  
SSRAS MLSS  
=QRAS  
=
As QWAS SSRAS « QMLSS and can be neglected  
Rearranging:  
 
!
QMLSS  
QRAS  
=
SSRAS MLSS  
11.5 Process trouble shooting  
The most common activated sludge quality issues include:  
1. Straggler floc: This condition shows small, light, and fluffy floc particles with poor settling char-  
acteristics. It occurs due to young sludge or low MCRT/MLSS levels and may be related to an  
inadequate microorganism population or an excessive BOD load (high F:M) which causes a log  
growth situation. The cells become dispersed rather than flocculated, settleability is poor, and the  
effluent becomes turbid. In this condition oxygen is used up quickly due to the high metabolism  
rate, and sludge production is high. One tell-tale sign of this condition is the production of huge  
amounts of a billowing white foam.  
2. Pin floc (Dispersed Growth): Here flocs - pin floc is seen in the effluent. These are larger dimension  
spherical particles. The sludge settles well in the settleability test but the supernatant is cloudy.  
This floc consist mostly of floc-forming bacteria without a filament backbone. This condition  
occurs at starvation condition when all of the influent BOD has been used up (low F:M) and the old  
sludge age organisms are now in endogenous respiration.  
3. Rising sludge: Under this condition, sludge settles well, compacts on the bottom of the clarifier,  
then starts to rise in clumps and patches to the surface. Rising sludge is typically evidenced as  
brown in color. Rising sludge occurs due to either denitrification or sludge septicity due to an  
excessive detention time in the clarifier.  
4. Bulking and foaming: Typical bacteria present in activated sludge include spherical, rod shaped  
 
11.5 Process trouble shooting  
171  
and filamentous. The presence of filamentous bacteria in activated sludge provide the important  
structural element for the bacterial floc. Bulking is indicated by SVI > 150 ml/g. However, certain  
condition cause an excessive growth of filaments which interferes with the settling (bulking) and  
cause foaming during aeration. Identifying which filaments are dominating in the system through  
a microscopic evaluation will help the operator to understand the condition in the treatment system  
so that corrective changes can be made.  
Filaments proliferation  
a. Bulking: Activated sludge bulking is marked by poor compaction – high SVI30 and poor solids-  
liquid separation evidenced by high effluent TSS. To resolve bulking issue, it is important to find  
its cause and take remedial actions. This is accomplished through a microscopic examination  
to identify the type of the filament to provide a clue on the cause. Short term remedy to bulking  
includes Chlorinating – 2-3 mg/l per 1000 lbs MLVSS or use of polymers.  
b. Foaming: Activated sludge foaming is caused mostly by two filaments: Nocardia spp. and Mi-  
crothrix parvicella (there are other non-filament causes of foaming - including excessive presence  
of surfactants. Both of these filaments have three causes in combination: (1) high grease and oil;  
(2) longer sludge age; and (3) low oxygen conditions or septicity. Foaming if left uncontrolled  
could potentially pose a permit compliance issue if the foam spills into the effluent. Additionally,  
Nocardia from the secondary sludge could cause digester foaming.  
Filament Type  
Cause  
Remedy  
Low aeration basin DO concentration for the applied F/M leads to filamen-  
tous bulking by several filaments. Typically a minimum DO of 2 mg/l is  
required for F:M upto 0.5. The condition may also be remedies by raising the  
MLSS concentration (decreasing the F/M)  
Low D.O. ( For the ap-  
plied organic loading )  
S.natans, Type 1701 and H.hydrossis  
Increase the F:M by reducing the MLSS concentration. When reduction in  
the MLSS concentration may not be feasible due to impact on nitrification,  
other suitable methods such as the use of a selector may be considered  
M.parvicella, Nocardia species, Type 0041,  
Type 0675, Type 1851 and Type 0803  
Low Organic Loading  
Rate ( Low F:M )  
Thiothrix I and II, Beggiatoa species, N.  
limicola II, Type 021N, Type 0092, Type  
0914, Type 0581, Type 0961 and Type 0411  
Septic Wastes / Sulfides (  
High organic acids )  
Reduce septicity using pre-aeration and through use of chemicals which  
prevent septicity and reduce sulfides  
Nutrient Deficiency ( N  
and / or P ) ( Industrial  
Wastes Only )  
Thiothrix I and II and Type 021N (N defe-  
ciency) N. limicola III (P defeciency)  
Nocardia species, M.parvicella and Type  
1863  
High Grease / Oil  
The aeration basin pH should be maintained in the range 6.5 to 8.5. Low pH,  
<6.5, may cause the growth of fungi and fungal bulking. The aeration basin  
pH can be adjusted using caustic, lime or magnesium hydroxide.  
Low pH ( Below pH 6.0 )  
/ Oil  
Fungi  
Table 11.1: Troubleshooting Activated sludge filaments  
11.6 Process modifications  
173  
11.6 Process modifications  
11.6.1 Conventional activated sludge process  
Process description  
The primary effluent (AS influent) - Q, along with the return activated sludge (RAS) - QRAS, are intro-  
duced at the head of the aeration tank. The oxygen demand is the highest at this point and decreases  
uniformly as the wastewater travels from one end of the tank to the other.  
Figure 11.1: Conventional Activated Sludge Process  
Process advantages and applications  
The conventional activated sludge process is most suitable for low-strength, domestic wastes with minimal  
peak load considerations as this process is susceptible to failure from shock loads. Shock Load is elevated  
strength (higher BOD and contaminant levels) in wastewater for brief period of time.  
Reasons for modifying the Conventional Activated Sludge Process  
Several factors contribute to the need for modifying the Conventional Activated Sludge Process:  
Unique characteristics of the influent flow which may include - high BOD loads, potential toxic  
loads, storm flows, loads with high carbonaceous BOD but low nitrogen  
• Space constraints  
• Energy and labor (O&M) constraints  
• Constraints requiring low solids production  
11.6.2 Conventional step feed process  
Process description  
In this process modification, the primary effluent is fed to the aeration tank at several different locations  
within the tank. The distribution of the influent along the length of the aeration reactor allows for distri-  
bution of the load (BOD) over the aeration tank and reduce oxygen sags in the aeration tank. The return  
sludge is introduced separately from the primary effluent and in many cases is allowed a short re-aeration  
period before mixing with primary effluent.  
Process advantages and applications  
Less aeration volume required to treat the same volume of wastewater compared to the conven-  
tional aeration.  
• Better control in handling shock loads  
• There is potential for:  
 
 
 
174  
Chapter 11. Activated Sludge  
Figure 11.2: Conventional step feed  
lower applied solids to the secondary clarifier  
lower oxygen requirement  
The step-feed process can be utilized on a variable basis using various modes (varying percent of  
the influent at the feed points) depending on the influent situation and the selection of the proper  
mode depends on the influent characteristics and the capabilities of the plant to handle the charac-  
teristics  
Figure 11.3: Step feed modification for Biological Nutrient Removal (BNR)  
11.6.3 Contact stabilization  
Process description  
• This process modification requires two aeration tanks.  
1. Return Sludge Re-aeration tank:  
This tank is for separate re-aeration of the return activated sludge before it is mixed with  
the incoming wastewater  
The residence time in the stabilization tank is typically 4 to 6 hours  
2. Mixed Liquor Aeration tank (or Contact tank):  
This tank is for mixing of the primary effluent with the re-aerated return activated sludge  
Here, the organisms quickly take in and store large amounts of food from the primary  
effluent.  
The residence time in this tank is approximately 30 to 90 minutes  
Process advantages and applications  
Because the contact stabilization system has a reservoir of microbes in the re-aeration tank, it is better  
able to withstand solids washouts that accompany major storms and avoid microbial kill-offs related to  
short-term toxic dumps  
 
11.6 Process modifications  
175  
Figure 11.4: Contact Stabilization  
11.6.4 Pure oxygen system  
Process description  
Very similar to conventional activated sludge except that high purity oxygen is supplied to aeration tank  
(reactor) rather than air and this process treats more wastewater in a smaller space in a shorter period of  
time. The Aeration tanks (reactors) are enclosed and sealed to prevent the oxygen from escaping from  
the tanks. Pure oxygen from an oxygen generation system or is introduced in the headspace of the reactor.  
Mechanical surface aerators agitate the mixed liquor (essentially splash the liquid in the headspace) to  
dissolve the oxygen into the mixed liquor. High DO levels are maintained in the aeration tanks (4 mg/l to  
15 mg/l). Roughly 90%-99% gaseous O2 goes in and 45-70% gaseous O2 is ventilated at the effluent end  
of the reactor.  
Figure 11.5: Pure Oxygen System  
Process advantages and applications  
• Pure oxygen systems are generally more stable and can treat large volumes of wastewater.  
• It treats more wastewater in a smaller space in a shorter period of time  
 
176  
Chapter 11. Activated Sludge  
11.6.5 Sequencing batch reactors  
Process description  
Rather than treating a continuous flow through a number of tanks, Sequencing Batch Reactors (SBRs)  
treat flows in “batches” through a single tank - "reactor" through sequencing stages. It performs the same  
treatment steps as a conventional AS plant, but they do it in sequenced stages rather than in a continuous  
flow stream. Most commonly preliminary treated wastewater is directly fed to the SBR.  
Figure 11.6: Sequencing Batch Reactor  
Typically SBR is operated in five stages:  
1. Filling: Influent raw wastewater enters the reactor  
2. Mixing & Aeration: This stage is where the BOD consumption and biomass production occurs.  
Mixing and aeration may be alternated to accomplish nitrification (during aeration) and denitrifica-  
tion (during mixing - without aeration - anoxic).  
3. Settling - This is the clarification stage. - to allow for the activated sludge floc to settle.  
4. Extraction - in this stage the supernatant treated wastewater is removed (for disposal) and part of  
the settled sludge is wasted.  
5. Idle - This stage is essentially to prepare the reactor and the biomass to receive the raw wastewater  
- Filling stage. In this stage some aeration/mixing may be provided. The biomass in the reactor  
enters the endogenous phase as the food is absent. The biomass metabolizes the stored food and the  
organisms feed on one another.  
 
11.6 Process modifications  
177  
Process advantages and applications  
• SBRs are ideal for small facilities as the process can be easily automated  
SBRs are very stable due to the high sludge ages maintained and the fact that all treatment takes  
place in a single tank  
SBR construction is less costly due to the elimination of secondary clarifiers and sometimes diges-  
tion facilities (fewer structures in general)  
11.6.6 Extended aeration  
Figure 11.7: Extended Aeration  
Process description  
This process typically directs raw sewage flow (without grit/screenings) into the aeration tank where it  
undergoes treatment for an extended period (usually 8 hours or more – 15 hours typical) followed by  
clarification. Mixed liquor concentrations are usually maintained at high concentrations. Typical F:M  
ratios are about 0.05 to 0.15 and MCRT of about 30 days.  
Process advantages and applications  
This process modification is appropriate for small facilities that have light solids loadings – plants  
typically treating less than 1 MGD – as the aeration tank volume requirement is larger  
In addition to consuming all the incoming food, the microbes consume all food stored within cells -  
endogenous respiration, and other dead microbes in the system  
• Extended aeration does not produce as much waste sludge as other process modifications  
The oxidation ditch is a variation of the extended aeration process. In the oxidation ditch, the aer-  
ation basin is ring or oval shaped. Aeration rotors or brushes cause the wastewater to flow around  
the ring and also aerate the wastewater.  
 
178  
Chapter 11. Activated Sludge  
Figure 11.8: Extended Aeration - Oxidation Ditch  
11.6.7 Krauss process  
Process description  
The process involves use of two aeration tanks. One aeration tank (Aeration Tank B in the graphic), is  
for blending return activated sludge (RAS) with anaerobic digester supernatant or sludge and aerating  
the solution. The other tank receives primary effluent (Aeration Tank A), return activated sludge, and the  
aerated mixture of return sludge and digester supernatant/sludge. Air (oxygen) is added to both aeration  
tanks.  
Figure 11.9: Krauss Process  
Process advantages and applications  
This process modification is widely used when wastewater contains a high ratio of carbonaceous to ni-  
trogenous material in the wastewater (nitrogen deficient). This typically occurs when there is significant  
contribution of wastewater from canneries or dairies. When the little nitrogen present in the influent  
wastewater is consumed, further consumption of carbonaceous material ceases. The digester supernatant  
or sludge being typically rich in nitrogenous material provides the required nitrogen for the nitrogen  
deficient process stream.  
 
Chapter Assessment  
1. What is the activated sludge floc made of  
2. What properties of activated sludge floc are key to the effectiveness of the activated sludge process  
3. List the activated sludge process control parameters (names/description)  
4. List the key design differences between a rectangular and circular primary clarifier  
5. Why is it important to ensure having a "good" microbiological composition of the activated sludge  
process (5 points)  
6. What is/are the main factor/s that control the microbial population  
7. MLVSS represents the  
fraction of the MLSS  
to  
8. Optimal range of SVI is between  
9. Straggler floc is associated with [type1] sludge while pin-floc is associated with  
Bulking and foaming is due to bacteria  
sludge  
10. Portion of the activated sludge floc settled in the clarifier that is returned to the front of the aeration  
basin to seed the incoming primary effluent is called [name]  
11. Which one of the following statements is TRUE regarding the various modifications of the acti-  
vated sludge process  
a. MCRT of 5 to 10 days is typical for extended aeration  
*b. Typical hydraulic detention times in the contact tank of the contact stabilization process need  
only be 0.5 to 1.0 hour.  
c. F to M ratios of 0.03 to .1 are appropriate for the step-aeration mode of the activated sludge  
process  
d. Pure oxygen activated sludge floc often has a large population of rotifers.  
e. Step feed -aeration involves decreasing the air being fed along the length of the aeration tank.  
12. What is the significance/importance of measuring OUR and SOUR and what are their respective  
units of measurement  
Correct Answer(s):  
13. List the advantages and disadvantages of the constant RAS flow control  
180  
Chapter 11. Activated Sludge  
Correct Answer(s):  
14. Name and describe the two RAS control approaches  
Answer the following related to activated sludge floc:  
15. What is the activated sludge floc made of (3 points)  
16. Activated sludge is an anaerobic process  
a. True  
*b. False  
17. Secondary treatment is mainly to remove the organic content of the wastewater  
*a. True  
b. False  
18. The contents of an aeration tank utilized in activated sludge treatment is referred to as mixed liquor.  
*a. True  
b. False  
19. In conventional activated sludge plants, six to eight hours of aeration detention time is used for  
acceptable plant operation.  
*a. True  
b. False  
20. Bulking occurs in primary clarifiers and is associated with improper scum removal.  
a. True  
*b. False  
21. Contact stabilization is a modification of the conventional activated sludge system.  
*a. True  
b. False  
22. In secondary settling tanks, the sludge pumping considerations would be the same as in primary  
settling tanks.  
a. True  
*b. False  
23. The main function of a launder in a secondary clarifier is to prevent scum and other floatables from  
leaving with the effluent flow  
a. T  
@Incorrect. Launder collects and conveys the effluent flow. Effluent baffles prevent scum and other  
floatables from leaving with the effluent flow  
*b. F  
@Correct. Launder collects and conveys the effluent flow. Effluent baffles prevent scum and other  
floatables from leaving with the effluent flow  
24. Excessive filamentous bacteria in activated sludge is typically controlled by bleach addition to RAS  
*a. True  
b. False  
25. SVI is a measure of the sludge volume that needs to be wasted  
a. True  
*b. False  
26. pH has little effect on the activated sludge plant  
a. True  
*b. False  
27. Bulking is caused by excessive filamentous bacteria  
11.6 Process modifications  
181  
*a. True  
b. False  
28. Extended aeration involves operating the activated sludge process at a high F: M ratio  
a. True  
*b. False  
29. In conventional secondary wastewater treatment processes, aerobic decomposition of solids will  
occur.  
*a. True  
b. False  
30. In the activated sludge process, the wastewater oxygen demand may be separated into two cate-  
gories: carbonaceous and nitrogenous  
*a. True  
b. False  
31. MCRT refers to the average number of days that a “cell” remains in an activated sludge system.  
*a. True  
b. False  
32. In activated sludge treatment a young sludge age is marked by a low F:M ratio  
a. True  
*b. False  
33. The "M" in the F:M ratio is the mass of mixed liquor suspended solids in the aeration basin  
a. True  
*b. False  
34. The SVI test is used for establishing amount of sludge to be wasted  
a. True  
*b. False  
35. A WAS or RAS flow change of 25% in one day will have little impact on the activated sludge  
treatment process  
a. True  
*b. False  
36. Rotifers are the dominant microorganisms in a young activated sludge  
a. True  
*b. False  
37. The F in the F to M ratio refers to the pounds of mixed liquor volatile suspended solids under  
aeration in an activated sludge plant.  
a. True  
*b. False  
38. The use of F:M ratio for controlling the activated sludge process implies the need for higher mass  
of microorganisms to treat a stream with a higher BOD  
*a. True  
b. False  
39. The white billowing foam commonly seen during the startup of the activated sludge plant is caused  
by low F:M ratio  
a. True  
*b. False  
40. When an activated sludge plant is first started, one should expect to see foaming  
182  
Chapter 11. Activated Sludge  
*a. True  
b. False  
41. MLVSS represents the  
42. Optimal range of SVI is between  
43. Straggler floc is associated with [  
sludge  
fraction of the MLSS  
sludge while pin-floc is associated with  
bacteria  
44. Bulking and foaming is due to  
45. Portion of the activated sludge floc settled in the clarifier that is returned to the front of the aeration  
basin to seed the incoming primary effluent is called [name]  
46. The basic objective in the activated sludge process is to maintain balanced conditions in the aera-  
tion basin, this balance is called:  
a. Endogenous respiration  
*b. Food/microorganism ratio  
c. Equilibrium status  
d. Mass balance ratio  
47. In the activated sludge treatment process, there are several control methods. One method is to  
maintain a BOD:MLVSS ratio. This is commonly referred to as:  
a. MCRT.  
b. SA.  
c. SA:SDI.  
*d. F:M.  
e. TS:SRT  
48. In calculating the detention time in an aeration tank, which one factor would not be considered?  
a. tank volume  
b. RAS flow  
c. plant flow  
*d. MLSS concentration  
e. none of the above  
49. The BOD loading rate divided by the quantity of microorganisms present in the biological reactors  
(aeration tanks) is known as:  
a. organic loading  
b. toxicity  
c. hydraulic loading  
*d. food to microorganism ration F:M  
50. An activated sludge process that has a desired F/M ratio of 0.05 and a sludge age of 30 days is what  
type of activated sludge process modification?  
*a. Extended aeration  
b. Conventional  
c. Complete mix  
d. Oxidation ditch  
51. Two major operational difficulties which sometimes occur in activated sludge secondary clarifiers  
are:  
*a. Low D.O. and algae growth  
b. Short circuiting and scum accumulation  
c. Rising sludge and bulking sludge  
11.6 Process modifications  
183  
d. Long detention time and short MCRT.  
52. A thick, scummy, dark tan foam on the surface of an activated sludge aeration tank is an indication  
of:  
*a. Aeration tank is underloaded (high MLSS.  
b. Aeration tank is overloaded (low MLSS.  
c. Excess grease in raw wastewater  
d. Excess phosphates (detergents. in raw wastewater  
53. A good quality of activated sludge is shown by:  
a. Black color and very small particle size  
b. Finely dispersed milky white particles  
c. A chocolate brown MLSS that does not settle well in the jar test  
d. A sludge that settles in one minute in the jar test  
*e. A chocolate color which settles out in 20-30 minutes with a D.O. of 2.0  
54. An aerobic treatment process is one that requires the presence of:  
a. Ozone  
b. organic oxygen  
c. no oxygen  
d. combined oxygen  
*e. dissolved oxygen  
55. An increasing F/M ratio and decreasing MCRT indicates  
*a. Excessive solids wasting causing a decrease in solids inventory  
b. Inadequate solids wasting causing an increase in the solids inventory  
c. Decreased hydraulic load increasing the sludge detention time  
d. Operation is normal  
56. A rapid and significant increase in filamentous organisms in the mixed liquor may be expected to:  
a. Result in a far better effluent because of the great amount of surface area for absorption  
b. Plug up the return sludge pumps because the filaments hang upon valves and gaskets in the  
sludge line  
c. Lead to much denser return sludge because the filaments would tend to strain the dispersed cells  
of ordinary organisms out of the effluent  
*d. Cause bulking of the sludge solids to the point that some solids might be swept out along with  
an otherwise clear liquid phase and result in turbid, poor quality effluent  
e. Lead to a much lower F/M ratio because the filaments are so totally insoluble.  
57. In the activated sludge treatment process, there are many control methods. One method is to main-  
tain a constant BODs:MLVSS ratio. sludge treatment process. This is commonly referred to as:  
a. MCRT  
b. SA  
c. SA: SDI  
*d. F:M  
e. TS:SRT  
58. The SVI of activated sludge is defined as:  
a. the volume of settled mixed liquor after 30 minutes or settling·  
b. the weight in grams of 200 ml of settled activated sludge  
*c. the volume in ml of 1 gram of activated sludge after 30 minutes of settling  
d. the total volume of MLSS in the aeration tank  
184  
Chapter 11. Activated Sludge  
e. the volume of settled sludge in the secondary clarifier  
59. The amount of air required in the operation of an activated sludge aeration tank is independent of  
the:  
a. temperature  
b. flow  
c. detention time  
d. organic loading  
*e. none of the above  
60. The successful operation of an activated sludge plant requires the maintenance of proper solids  
concentration in the system. One major limiting factor is:  
a. mixed liquor tank volume  
b. effluent flow  
*c. air supply  
d. chlorine demand  
e. none of the above  
61. The main difference between primary and secondary clarifiers is the:  
a. overall dimensions  
b. type of outlet weirs  
*c. density of sludge  
d. detention period  
e. flow distribution  
62. Given the data below, what is the most likely cause of the extended aeration facility problem?  
DATA: DO level high  
Blower normal  
Wastewater characteristics normal  
Drop pipe air control valves open  
Surface turbulence high  
a. Air relief valve stuck shut  
*b. Blower speed too fast  
c. Blower speed too slow  
d. Drop pipe air control valves not open far enough  
63. What test is used to determine the organic matter found in the mixed liquor?  
a. COD  
b. MLSS  
*c. MLVSS  
d. TOC  
64. Fixed porous plate diffusers can be cleaned by scrubbing with  
a. Detergent  
b. A strong acid solution  
*c. A strong chlorine solution  
d. A weak sodium hydroxide solution  
65. A 30 minute settleability test MLSS sample should be collected:  
a. At the primary clarifier effluent  
b. In the return sludge line  
c. Where the return sludge mixes with the aeration basin contents  
11.6 Process modifications  
185  
d. At the aeration basin influent  
*e. At the aeration basin outlet  
66. A consulting engineer has recommended addition of a roughing filter and intermediate clarifier  
between your primary clarifier and aeration basin to better handle increasing industrial loads. This  
addition would:  
a. Be the best form of flow equalization available  
b. Remove most of the fixed dissolved solids  
c. Reduce drastically the fine dissolved matter  
d. Cost a lot and do nothing  
*e. Reduce the organic load on the aeration basin  
67. An activated sludge process that has a desired F/M ratio of 0.05 and a sludge age of 30 days is what  
type of activated sludge process modification?  
*a. Extended aeration  
b. Conventional  
c. Complete mix  
d. Oxidation ditch  
68. Two major operational difficulties which sometimes occur in activated sludge secondary clarifiers  
are:  
*a. Low D.O. and algae growth  
b. Short circuiting and scum accumulation  
c. Rising sludge and bulking sludge  
d. Long detention time and short MCRT.  
69. Two major operational difficulties which sometimes occur in activated sludge secondary clarifiers  
are:  
*a. Low D.O. and algae growth  
b. Short circuiting and scum accumulation  
c. Rising sludge and bulking sludge  
d. Long detention time and short MCRT.  
70. Possible techniques for controlling filamentous organisms in an activated sludge process include:  
*a. Dosage of return sludge with a disinfectant such as chlorine or hypochlorite  
b. Lower DO levels in aeration bans so filamentous organisms cannot breathe or respire  
c. Lower F/M level to starve filamentous organisms  
d. Stop wasting to allow activated sludge bugs to gain control  
71. Possible techniques for controlling filamentous organisms in an activated sludge process include:  
*a. Dosage of return sludge with a disinfectant such as chlorine or hypochlorite  
b. Lower DO levels in aeration bans so filamentous organisms cannot breathe or respire  
c. Lower F/M level to starve filamentous organisms  
d. Stop wasting to allow activated sludge bugs to gain control  
72. An activated sludge process that has a desired F/M ratio of 0.05 and a sludge age of 30 days is what  
type of activated sludge process modification?  
*a. Extended aeration  
b. Conventional  
c. Complete mix  
d. Oxidation ditch  
73. An aerobic treatment process is one that requires the presence of:  
186  
Chapter 11. Activated Sludge  
a. Ozone  
b. organic oxygen  
c. no oxygen  
d. combined oxygen  
*e. dissolved oxygen  
74. An increasing F/M ratio and decreasing MCRT indicates  
*a. Excessive solids wasting causing a decrease in solids inventory  
b. Inadequate solids wasting causing an increase in the solids inventory  
c. Decreased hydraulic load increasing the sludge detention time  
d. Operation is normal  
75. A rapid and significant increase in filamentous organisms in the mixed liquor may be expected to:  
a. Result in a far better effluent because of the great amount of surface area for absorption  
b. Plug up the return sludge pumps because the filaments hang upon valves and gaskets in the  
sludge line  
c. Lead to much denser return sludge because the filaments would tend to strain the dispersed cells  
of ordinary organisms out of the effluent  
*d. Cause bulking of the sludge solids to the point that some solids might be swept out along with  
an otherwise clear liquid phase and result in turbid, poor quality effluent  
e. Lead to a much lower F/M ratio because the filaments are so totally insoluble.  
76. During severe cold weather operation of an activated sludge plant biological activity and clarifier  
sludge settling is reduced. White of the following might help?  
*a. Increase the MLSS  
b. Decrease the MLSS  
c. Increase the D.0.  
d. Decrease the D.0.  
e. Add ammonia  
77. Excess white foam in an aeration basin can be corrected by  
a. Decreasing the aeration rate  
b. Decreasing detention time  
*c. Increasing the MLSS  
d. Decreasing the MLSS  
e. Increasing aeration rate  
78. Given the following data, what is the most likely cause of the activated sludge problem?  
DATA: The aeration tanks in an activated sludge plant have maintained a stable white foam with a  
brownish tint less than one inch thick. ·  
BOD removals have been at their normal high efficiency.  
Settling of the activated sludge in the secondary clarifiers has been good - as is normal.  
Air supplied to the system has been a normal 30,000 cfm, with a consistent DO of 2.5 mg/L.  
MLSS has been maintained at 2,500 mg/L - normal.  
Gradually during your shift the DO has risen to 5.0 mg/L.  
a. A toxic substance has affected the activated sludge.  
b. BOD loading on the aeration system has increased.  
c. Increased BOD loading has caused a corresponding increase in activated sludge activity.  
*d. No change.  
79. Given the following data, what is the most likely cause of the secondary sedimentation tank prob-  
11.6 Process modifications  
187  
lem?  
DATA: Sludge depth in tank too high.  
Tank effluent turbid.  
Tank effluent requiring above normal chlorine dosage. Sweeparms in tank bottom operating.  
Return activated sludge flow to aeration tank low.  
Controls on return activated sludge pump on automatic.  
Control sensors for return sludge operating normally.  
a. Accuracy of sludge depth measurement.  
b. Return activated sludge pump worn, needing repair.  
*c. Speed of sweeparms travel.  
d. Sweep arm overload tripped.  
80. How many gallons of paint will be required to paint the walls of a 40 ft long x 65 ft wide x 20 ft  
high tank if the paint coverage is 150 sq. ft per gallon. Note: We are painting walls only. Disregard  
the floor and roof areas.  
*a. 28 gallons  
b. 63 gallons  
c. 35 gallons  
d. 56 gallons  
81. If there is an insufficient supply of air or oxygen being introduced into the aeration tank of an  
extended aeration plant, the liquid in the tank will likely  
a. Contain a very fine light brown floc.  
b. Contain very small air bubbles.  
*c. Have a black or blackish appearance and an offensive odor.  
d. Have a dishwater appearance and a greasy odor.  
82. If the return sludge pump does not function the effect on other unit processes will be to:  
a. Tum the aeration basin influent dark  
b. Increase chlorine residual  
*c. Increase effluent suspended solids  
d. All the above  
e. None of the above.  
83. If the sludge depth in a secondary sedimentation tank is too high, what will happen?  
a. Decreased turbidity in effluent.  
b. Return activated sludge will have lower oxygen demand.  
c. Settleable solids from aeration tank will increase.  
*d. Sludge may become septic.  
84. If you must waste sludge from an activated sludge plant the maximum rate is:  
*a. 20 % per day  
b. 40 % per day  
c. 60 % per day  
d. 75% per day  
e. 100 % perday  
85. In an activated sludge system, what is perhaps the most important parameter affecting biological  
activity?  
a. pH.  
b. Alkalinity.  
188  
Chapter 11. Activated Sludge  
*c. Dissolved oxygen.  
d. Temperature.  
86. Mean cell residence time (MCRT. represents the theoretical time that a microorganism stays in the  
activated sludge system. The typical values for most activated sludge processes are:  
a. 3 - 30 days  
*b. 3 - 15 days  
c. 5 - 15 days  
d. 5 - 20 days  
87. Nocardia is associated with a particular type of brown, viscous scum or foam on the surface of  
the activated sludge aeration tank. One operational strategy that has been somewhat successful in  
reducing the severity of this foam is:  
a. to increase the plant’s MCRT.  
*b. to decrease the plant’s mixed liquor concentration.  
c. to operate at an F:M ratio of less than 0.025.  
d. to spray the foam with fine mist water sprays.  
e. to increase the luxury DO concentration at the end of the aeration tank.  
88. One limitation in using constant mixed liquor volatile suspended solids (MLVSS) or mixed liquor  
total suspended solids (MLTSS) as the control methodology for activated sludge treatment is  
*a. In practice it is not possible to operate at a constant MLTSS or MLVSS.  
b. It is based on consistency of raw waste load which seldom exists.  
c. Most facilities don’t have the lab equipment necessary to determine MLVSS.  
d. None of the above.  
89. Define Nocardia & problems associated with it. List 5 methods of controlling Nocardia.  
Response:  
a Define Nocardia & problems associated with it.  
Nocardia is a type of filamentous organism which on overgrowth in the mixed liquor is  
the cause of foaming during aeration in the activated sludge process. Nocardia prolifer-  
ation is associated with the following three causes in combination: (1) high grease and  
oil; (2) longer sludge age; and (3) low oxygen conditions or septicity.  
Nocardia develops a persistent, viscous brown foam scum. If not skimmed properly,  
this can cause an increase in both suspended solids and BOD and cause scum spill over  
on to the catwalks and make them sticky and slimy. Additionally, Nocardia from the  
secondary sludge could cause digester foaming.  
b List 5 methods of controlling Nocardia.  
i. Reduce sludge age by increasing wasting rate. This would also effectively reduce MLSS  
concentration in the aeration tank and increase F:M.  
ii. Mechanically skim and remove the foam.  
iii. Enforce industrial waste control program and manage the collections cleaning proce-  
dures to minimize the amount of grease and oil in the influent wastewater.  
iv. Chlorinate the RAS stream or the mixed liquor return or by spraying directly into the  
aeration tank.  
v. Use of mannich polymer to enhance settling and removal of the Nocardia filaments  
11.6 Process modifications  
189  
90. Clearly show how SOUR & OUR are different. List two possible causes for a sudden 40% decrease  
in OUR during the last 12 hours of operation. What lab test might confirm or refute the two causes  
of this sudden decrease in OUR. Response:  
a Clearly show how SOUR & OUR are different.  
Oxygen Uptake Rate (OUR) involves measurement of the amount of oxygen used up by  
the microorganisms in the mixed liquor using a DO probe and is expressed in unit time  
of mg/L-hr (ppm O2 consumed per hour). By knowing the OUR, we can establish the  
activity of the microorganisms in the aeration tank and know if they are consuming the  
oxygen provided for removing organic matter. For conventional activated sludge process  
the typical OUR values range from 10 to 30 mg/L-hr  
Specific Oxygen Uptake Rate (SOUR) provides the OUR information based on the  
concentration of microorganisms present. SOUR is obtained by dividing OUR with  
MLVSS.  
mg  
O
2
mg  
gm  
lhr  
The value is indicated and measured in terms of unit of  
1000  
.
mg  
l
MLVSS  
Optimal range of SOUR is usually between 8 to 20.  
b
List two possible causes for a sudden 40% decrease in OUR during the last 12 hours of  
operation.  
i. Presence of toxic substance in the mixed liquor inhibiting the normal biological activity  
in the aeration basin  
ii. Excessive loss of biomass due to inadvertent wasting  
c What lab test might confirm or refute the two causes of this sudden decrease in OUR.  
i. Run a MLSS test to see if adequate biomass is present  
ii. Run a COD test to ensure adequate amount of organics (food) is present  
(If there is sufficient biomass and food present, the only reason why the OUR would  
have declined would be because of toxic substance entering the system)  
91. The control and calculation of RAS flow rates are important considerations in the Operation of  
activated sludge wastewater treatment plants. Most commonly the operator Either sets the return  
rate (Qr) at a constant flow or as a constant percentage of flow.  
(a) Identify two effects that each of these approaches will have on plant operations.  
Assume that a normal diurnal variation in both wastewater flow and strength.  
(b) The so-called “solids balance approach” (shown below) may be used to Mathematically esti-  
mate Qr. What two assumptions are made in deriving this formula?  
[Qr + Q] x MLSS Conc. = Qr x RAS Conc.  
(c) Using the information given below and the “solids balance” equation, calculate SVI and Qr  
comment on these values:  
Q. ... . . . . . . . . . . . . . . . . . ..2.0 MGD  
MLSS. . . . . . ..... . . . . 2350 mg/L  
RAS.. . . . . . . . . . . .. . . .7350 mg/L  
SV30.. . . . . . . . . . . . . . . . . 320 mLs/g  
12. Nutrient Removal  
12.1 Importance of removing nutrients  
Plant nutrients - nitrogen and phosphorous, if present in wastewater effluent discharge promote  
growth of plant and algal matter in the receiving waters causing destruction of the normal aquatic  
life mainly due to oxygen depletion - eutrophication.  
Because of the potential impacts of the presence of these nutrients in wastewater effluent on the re-  
ceiving waters, limits on the levels of these nutrients is typically stipulated in the treatment plant’s  
wastewater discharge permit.  
Typically, conventional secondary treatment processes are designed primarily remove the organics  
from the wastewater. Secondary treatment process designed to additionally remove nutrients is  
deemed as tertiary or advanced treatment is termed as Biological Nutrient Removal (BNR).  
12.2 Nitrogen Removal  
Nitrogen exists in many forms and changes from one form to another in the environment as part of the  
nitrogen cycle below.  
In raw domestic wastewater, nitrogen exists primarily as organic nitrogen (40%) and ammonia  
nitrogen (60%).  
The sum of organic nitrogen and ammonia nitrogen is referred to as “Total Kjeldahl Nitrogen”  
(TKN).  
The important forms of nitrogen as part of wastewater treatment are: N2 (Nitrogen Gas), NO2  
(Nitrite), NO3 (Nitrate), NH3 (Ammonia), NH4 + (Ammonium Ion), and Organic Nitrogen.  
Typical concentrations of the nitrogen constituents in wastewater are tabulated below.  
Ammonia (NH3) and ammonium (NH+4 ) are both commonly referred to as “Ammonia-nitrogen (NH3-N)”  
These two forms of nitrogen can rapidly change from one to the other depending on pH and temperature.  
 
 
 
192  
Chapter 12. Nutrient Removal  
Figure 12.1: Nitrogen Cycle  
The figure below illustrates the relative distribution of ammonia (unionized ammonia) and the ammonium  
ion (ionized ammonia)in water as a function of pH.  
In biological treatment processes, the organic nitrogen is quickly changed to ammonia nitrogen by  
natural processes.  
Ammonia-nitrogen is toxic to aquatic life and toxicity is affected by both, temperature and pH of  
water.  
Toxicity of ammonia nitrogen increases as temperature increases  
Ammonia nitrogen is toxic when present as unionized NH3. This unionized ammonia NH3  
is the predominant form of ammonia nitrogen at higher pH. The ammonia nitrogen (NH3-N)  
toxicity is lower when present as ionized ammonia (NH+4 )  
Juvenile fish is more susceptible to ammonia toxicity. Therefore, ammonia toxicity effects  
will may be more pronounced during the fish breeding season in spring  
12.2.1 Nitrogen Removal Methods  
Nitrification and denitrification  
This is the most common ammonia removal process and is accomplished during secondary treatment. It is  
a two-step process:  
1. Nitrification This involves conversion of ammonia to nitrate  
2. Denitrification This involves conversion of nitrate to gaseous nitrogen which escapes from the  
waster.  
NITRIFICATION  
Nitrification is a two step process:  
Step 1: Nitrosomonas and other similar bacteria oxidizes ammonium to nitrite via hydroxylamine.  
2NH4+ +O2 2NH2OH +2H+  
 
12.2 Nitrogen Removal  
193  
NITROGEN IN WASTEWATER  
Forms of Nitrogen  
Formula  
Found in  
Typical  
Concentration  
+
Ammonia/Ammonium  
NH3/NH4  
Influent wastewa-  
ter  
30-50 mg/l  
30-60 mg/l  
1-40 mg/l  
Total Kjeldahl Nitrogen  
(Ammonia/Ammonium + Organic Nitrogen)  
TKN  
TIN  
Wastewater  
effluent  
Total Inorganic Nitrogen  
(Ammonia/Ammonium + Nitrite + Nitrate)  
Wastewater  
effluent  
Nitrate  
Nitrate  
NO3 −  
Nitrified effluent  
1-35 mg/l  
0.1-2 mg/l  
NO2  
Partially nitrified  
effluent  
Table 12.1: Forms of Nitrogen in Wastewater  
Figure 12.2: Ammonia-Ammonium Equilibrium  
194  
Chapter 12. Nutrient Removal  
NH4+ +1.5O2 NO2 +2H+ +H2O  
Step 2: Conversion of nitrite to nitrate by nitrobacter and other similar bacteria  
NO2 +0.5O2 NO3  
The above two reactions are generally coupled and precede rapidly to the nitrate form; therefore,  
nitrite levels at any given time are usually below 0.5 mg/L.  
Nitrifiers (responsible for the consumption of nBOD) compete with the organotrophs which con-  
sume the cBOD in the secondary reactor.  
Nitrification occurs at the tail end of the secondary treatment reactor when the cBOD is low and the  
organotrophs are not as active  
• Nitrification occurs 3-4 times slower than carbonaceous oxidation  
For nitrosomonas to generate 1 part of new cells it needs 30 parts of ammonium ions and nitrobac-  
ter needs 100 parts of nitrite ions to generate 1 part new cells. Due to the high demand of the nitro-  
gen based ions, the reproductive rates of nitrifiers are very low thus plant upsets can take nitrifiers  
weeks to recover as opposed to days or hours for carbon bacteria.  
For each 1-gram of NH3-N oxidized to NO3, 0.15 grams of new bacteria cells are formed. Sludge  
generation from nitrifiers is minimal in a wastewater treatment plant.  
Controlling Factors for Nitrification:  
1. Dissolved oxygen Oxygen levels are critical for nitrification. Higher levels of oxygen are required  
for nitrification than for carbon removal. While 1 part of O2 is needed for removing 1 part of  
cBOD, 4.5 parts of O2 are needed for every part of NH4 +N (nBOD) to be degraded. In order  
for nitrification to occur, dissolved oxygen levels of 1.0 to 4.0 mg/L are usually maintained in  
the aeration tanks. Maximum nitrification occurs at DO levels of about 3.0 mg/L. Nitrification  
will therefore result in significantly higher power cost due to the needs for maintaining the higher  
oxygen level  
2. Alkalinity and pH  
Nitrification leads to a depletion of alkalinity in the activated sludge mixed liquor. Nitro-  
somonas and Nitrobacter use carbonate as their carbon (food) source and the ammonia is just their  
energy transfer source. As the total alkalinity of wastewater is due to the individual contributions  
of carbonate, bicarbonate and hydroxide ions, consumption of carbonate and bicarbonate ions by  
the nitrifiers results in the depletion of alkalinity. Alkalinity provides a buffer for pH change. If  
the alkalinity is reduced beyond certain levels, further formation of acidic metabolic byproducts of  
bacterial activity may lead the pH to decline inhibiting bacterial activity. 7.14 parts of alkalinity  
are required for each part of ammonia removed. Nitrification rates are rapidly depressed as  
the pH is reduced below 7.0. pH levels of 7.5 to 8.5 are considered optimal. That is why alkalinity  
is extremely critical in nitrification. An alkalinity of 60 mg/L in the secondary treatment reactor  
(aeration tank, trickling filter, RBC, etc.) is generally required to ensure adequate buffering  
3. MCRT, F/M, or Sludge Age For nitrification to occur the activated sludge treatment process  
needs to be operated at higher MCRT as the reproductive rates of nitrfiers is low. An MCRT  
of greater than 8 days is typically considered essential for nitrification to occur.  
4. Wastewater temperature Nitrification is inhibited at lower wastewater temperatures in wastew-  
ater treatment plants. To achieve the same level of nitrification, longer detention time may be  
needed in the winter versus the summer months since the activity drops significantly. Lower tem-  
perature effects on Nitrification may be partially mitigated by increasing MLVSS and MCRT. The  
12.2 Nitrogen Removal  
195  
Figure 12.3: Weather Effects on MCRT Requirements for Nitrification  
optimal temperature range for nitrification is between 60to 95degrees F. Below 40nitrification  
will probably not occur.  
5. Inhibition to nitrification by toxic compounds Many compounds can be toxic to nitrifiers- Nitro-  
somonas and Nitrobacter. These include unionized ammonia, heavy metals, solvents and cyanide.  
196  
Chapter 12. Nutrient Removal  
DENITRIFICATION  
Dentrification is a microbiological process in which the nitrate formed during nitrification is re-  
duced by bacteria in anoxic conditions to molecular nitrogen which is insoluble in water and es-  
capes into the atmosphere.  
A large percentage of the bacteria in the activated sludge process are capable of denitrification.  
In the absence of free molecular oxygen – under anoxic conditions, these bacteria present in the  
activated sludge process utilize the nitrate and nitrite to degrade the cBOD.  
NO3 +cBOD NO2+CO2 +H2O  
NO2 +cBOD N2 +CO2 +H2O  
• Approximately 50% of the alkalinity lost during nitrification is recovered during denitrification.  
The quantity of cBOD available is critical for denitrification to occur. Complete denitrification  
usually occurs at a cBOD to nitrate and nitrite ratio of approximately 3:1. Given the need for the  
presence of adequate cBOD for the denitrification step, a cBOD supplement (a suitable organic  
compound - example: alcohol) is utilized in certain configurations.  
If the aeration reactor effluent is nitrified, denitrification can occur in clarifier. The absence of  
aeration in the clarifier creates anoxic conditions which makes the condition favorable for denitri-  
fication to occur. Denitrification in the secondary clarifier is not desirable as the nitrogen evolved  
will cause the sludge settling in the clarifier to rise to the surface.  
Three of the more common methods to denitrify the nitrified effluent are:  
Method 1.The aeration tank is followed by a denitrifying tank which is not aerated but is supplied with  
organic substrate (cBOD source) such as methanol or acetic acid.  
Figure 12.4: Denitrification Method 1:  
Organic substrate addition  
Method 2.The mixed liquor along with RAS is recirculated to an anoxic zone at the inlet of the aera-  
tion tank where it is mixed with the influent. The anoxic zone has mixing provided but no aeration. The  
influent flow provides the necessary cBOD for the denitrification process.  
12.2 Nitrogen Removal  
197  
Figure 12.5: Denitrification Method 2:  
Mixed liquor recirculation  
Method 3.The aeration tank consists of alternating oxic (aerated) and anoxic zones  
Figure 12.6: Denitrification Method 3:  
Alternating oxic-anoxic zones  
Breakpoint chlorination  
• Breakpoint chlorination is a means of eliminating ammonia using chlorine  
As chlorine is added to water containing ammonia, it converts ammonia into chloramines which in  
the presence of additional free chlorine forms nitrogen gas which is released to the atmosphere  
Chlorination of a water containing ammonia results in the following:  
• an initial increase in combined chlorine residual  
• followed by a decrease in the combined chlorine residual along with ammonia concentrations  
followed by an increase in free chlorine residual and near complete removal of ammonia as nitro-  
gen gas.  
• Chlorine reacts with ammonia to form chloramines  
1. First the free chlorine in contact with ammonia forms monochloramine and water  
Monochloramine has disinfection properties  
Dominates when Cl:N mass ratio is 0 to 5:1  
The breakpoint curve rises at about 1:1 during monochloramine formation  
NH3 +HOCl NH2Cl(monochloramine)+H2O  
198  
Chapter 12. Nutrient Removal  
2. Monochloramine reacts further with chlorine to give dichloramine and water  
HOCl +NH2Cl NHCl2 +H2O  
Also, monochloramine auto decomposes into dichloramine  
2NH2Cl NHCl2 +NH3  
Between dichloramine is formed between 5:1 and 7:1 Cl:N mass ratio  
When you are getting significant dichloramine, the breakpoint curve will start dropping  
NH2Cl +HOCl NHCl2(dichloramine)  
at pH  
> 7.5, monochloramine is the dominant chloramine species as pH decreases from 7.5,  
dichloramine becomes the dominant chloramine species increases in the chlorine to nitrogen  
dose ratio results in corresponding increases of nitrogen trichloride, but only when the pH is  
< 7.4  
3. Formation of nitrogen trichloride from the reaction of chlorine and dichloramine does not  
typically occur as it is the favored product at low pH - <4  
NHCl2 +HOCl NCl3(nitrogentrichloride)  
4. Additional free chlorine +chloramines H+ +H2O+N2  
Chloramines have disinfection properties albeit much lower than free chlorine ( 5% of free avail-  
able chlorine) but last much longer in the system than free chlorine.  
After the breakpoint, free chlorine residuals develop. Free chlorine residuals usually destroy odors,  
kill microorganisms and oxidize organic matter.  
Breakpoint chlorination is the application of sufficient chlorine beyond the chlorine demand to  
maintain a free available chlorine residual.  
Theoretically chlorine requirement = Wt. NH3-N x 7.6  
in practice (Margin of safety) = Wt. NH3-N x 10  
Thus, breakpoint chlorination is possible if ratio of Cl2 to ammonia exceeds 10:1 then free Cl2  
may exist. Cl2 demand will be high because the reaction of free Cl2 with nitrite and other organic  
compounds.  
Theoretically, while microorganisms are killed as the chlorine demand is being satisfied, disin-  
fection is generally the result of chlorine residual or the amount of chlorine remaining after the  
chlorine demand has been satisfied.  
The following chart is a graphical explaination of the concept of Breakpoint Chlorination  
12.2 Nitrogen Removal  
199  
• Point A is at the beginning of chlorine application  
Between Points A and B, the chlorine dosage produces no residual because of an immediate chlo-  
rine demand caused by fast-reacting ions from metal salts and H2S.  
• Point B is the beginning of the reaction between chlorine and ammonia present  
• Mono and dichloramines are formed between points B and C  
Zone 1 - between points A and C, is the combined zone and has mono and di chloramines and  
ammonia. Mono chloramines is a stable disinfectant while dichloramines is a strong disinfectant  
but unstable.  
After the maximum combined residual is reached (point C), further chlorine doses decrease the  
residual due to chloramine oxidation to dichloramine, occurring between points C and D. This is  
Zone 2 - Breakpoint Zone  
Point D represents the breakpoint - the point at which chlorine demand has been satisfied and  
additional chlorine appears as free residuals  
Between points D and E, free available residual chlorine increases in direct proportion to the  
amount of chlorine applied. This is Zone 3 which is the free chlorine zone and has hypochlorous  
acid but no ammonia.  
Factors that affect breakpoint chlorination are initial ammonia nitrogen concentration, pH, tempera-  
ture, and demand exerted by other inorganic and organic species  
Weight ratio of chlorine applied to initial ammonia nitrogen must be 8:1 or greater for the break-  
point to be reached. If the weight ratio is less than 8:1, there is insufficient chlorine present to  
oxidize the chlorinated nitrogen compounds initially formed  
200  
Chapter 12. Nutrient Removal  
When instantaneous chlorine residuals are required, the chlorine needed to provide free available  
chlorine residuals may be 20 or more times the quantity of ammonia present. Reaction rates are  
fastest at pH 7-8 and high temperatures  
12.3 Phosphorous removal  
Limits related to phosphorous concentrations are commonly found in wastewater treatment facili-  
ties’ NPDES discharge permits.  
Phosphorous removal as part of the treatmnent process accomplishes meeting the NPDES require-  
ment also allows for its recovery for agricultural use.  
• Phosphorous is very reactive and occurs in nature as phosphate (PO4 ).  
In wastewater phosphorous is present mainly in an inorganic form as orthophosphate (typically  
about 10 ppm) and as organic phosphate (typically 6 mg/l).  
The inorganic phosphate is contributed by detergents and household cleaning products such as  
soaps  
Organic phosphate is contributed by human wastes and food residues as phosphate based molecules  
play a key role in biological (plant and animals) energy and lifecycles and present in sugars, phos-  
pholipids, and nucleotides. The organic phosphates are decomposed to orthophosphate.  
12.3.1 Phosphorous removal methods  
Chemical precipitation method  
In the chemical based phosphorous removal method, aluminum and iron salts such as ferric/ferrous chlo-  
ride and aluminum sulfate react with soluble phosphate to form solid precipitates that are removed by  
solids separation processes including clarification and filtration.  
Typically, the phosphate precipitant is applied in both the primary clarifier feed and also just before the  
secondary clarifier.  
Figure 12.7: Chemical removal of phosphorous  
Enhanced biological phosphorus removal (EBPR)  
• For the removal of phosphorous as part of the activated sludge treatment process:  
 
 
12.3 Phosphorous removal  
201  
The process is designed to promote the growth of specific aerobic microorganisms generally  
known as phosphorus accumulating organisms (PAOs) the activated sludge process.  
Examples of these bacteria in activated sludge: Acinetobacter, Rhodocyclus  
These bacteria have the ability to store substantial quantities of intracellular (inside the  
bacterial cell) phosphorus.  
*
*
PAOs maybe upto 40% phosphorous by weight.  
*
The process is operated by providing alternating anaerobic and aerobic conditions in the  
aeration basin of the activated sludge treatment process.  
EBPR process description:  
In the anerobic part of the aeration basin, the aerobic bacteria produce VFAs by breaking down  
organic matter in the influent wastewater. Thus, the readily biodegradable material - VFAs, is  
amply available for the PAOs when in the anaerobic environment.  
Here the PAOs consume the volatile fatty acids (VFAs) produced by the anaerobic fermen-  
tation of the organic matter in the anaerobic zone. Note: This is a fermentation process not  
anaerobic digestion (in digestion VFAs are destroyed and converted to methane).  
The VFAs are converted into energy rich, carbon based, storable food source.  
The energy for this conversion of the VFA into storable food comes from the breakdown of  
the PAOs polyphosphate reserves.  
Ortho phosphate is released during this phase.  
After exposure to enough VFA, the PAOs energy reserves will be depleted and become  
stressed.  
In the following aerobic phase of the process the PAOs multiply and take up phosphate to replenish  
the supplies depleted in the anaerobic phase.  
By oxidizing the carbon reserves built up in the anaerobic phase, PAOs are able to store more  
phosphate under aerobic conditions than was released under anaerobic conditions because  
considerably more energy is produced by aerobic oxidation of the stored carbon compounds  
than was used to store them under anaerobic conditions.  
Eventhough the EPBR requires only a good aeration, its success in removing phosphorous is dependent on  
the presence of adequate quantities of readily degradable organic material in the secondary influent.  
Chapter Assessment  
1. Nitrification is conversion of  
to  
2. Nitrosomonas are  
Correct Answer(s):  
a. Autotrophic  
(type) bacteria  
3. Denitrification requires  
Correct Answer(s):  
a. anoxic  
condition  
4. Presence of adequate quantity of  
Correct Answer(s):  
a. vfa  
is critical for phosphorous removal  
5. In nitrification,  
Correct Answer(s):  
a. 4.57  
parts of oxygen required per part of ammonium nitrogen  
6. In domestic water  
Correct Answer(s):  
a. 40  
7. Nitrification is conversion of  
to  
8. In domestic water  
Correct Answer(s):  
a. 40  
% of nitrogen is present as ammonia/ammonium  
9. The approximate pH for good nitrification is closest to  
a. 6.5  
b. 7.0  
*c. 8.5  
d. 10.0  
10. Which of the following biological processes can produce alkalinity?  
204  
Chapter 12. Nutrient Removal  
a. Carbonaceous BOD removal  
*b. Denitrification  
c. Nitrification  
d. Phosphorus removal by chemical addition with ferrous chloride  
11. In a typical biological nitrification system, an oxygen demand of  
parts oxygen per part of  
ammonia oxidized is exerted in the nitrification process.  
a. 2.37  
b. 5.47  
c. 27  
*d. 4.57  
12. The biological nitrification process is carried out by bacteria that convert ammonia nitrogen to  
nitrate nitrogen. the two (2) specific groups of bacteria that perform this conversion are:  
*a. Nitrosomonas and nitrobacter  
b. Flagellates and swimming ciliates  
c. Flagellates and crawling ciliates  
d. Crawling ciliates and swimming ciliates  
13. Biological denitrification requires  
a. Aerobic  
conditions.  
*b. Anoxic  
c. Anaerobic  
d. Facultative  
14. Denitrification is an activated sludge plant involves:  
a. Oxidation of ammonia to nitrate.  
b. High concentrations of D.O. in the mixed liquor as it settle in the secondary clarifier.  
c. Biological oxidation of nitrate to nitric oxide.  
*d. Biological reduction of nitrate to nitrogen gas.  
e. Poor compaction of mixed liquor as it settles in the secondary clarifier.  
15. In a nitrifying activated sludge plant, it is important to maintain adequate D.O. in the mixed liquor  
as it settles in the final clarifier to prevent:  
a. Death of organisms in the floc that “eat” organic materials.  
*b. Denitrification.  
c. Bulking.  
d. Growth of filamentous organisms.  
e. Death of filamentous organisms.  
16. Nitrification in activated sludge does not involve:  
a. The oxidation of ammonia to nitrite.  
*b. Removal of oxygen from nitrate to form nitrogen gas.  
c. The biological oxidation of nitrite to nitrate.  
d. Higher concentration of D.O. in the aeration basin.  
e. Control of alkalinity to stabilize pH.  
17. Limits on the concentration total NH3 - N are frequently placed on secondary effluents discharged  
into sensitive receiving waters. An important consideration in the setting of these limits by the  
RWQCB would be:  
a. The concentration of algal cells in the receiving water.  
*b. The pH & temperature of the receiving water.  
12.3 Phosphorous removal  
205  
c. The salinity of the receiving water.  
d. The population of nitrosomonas bacteria in the final effluent.  
e. The concentration of copper ions in the final effluent because these ions form a non-toxic com-  
plex with the ammonia nitrogen.  
18. Many NPDES permits limit the amount of ammonia nitrogen that can be discharged into a stream  
because:  
a. Ammonia exerts an oxygen demand in the stream  
b. Ammonia can be toxic to aquatic life  
c. Ammonia reacts with chlorine which can interfere with disinfection  
*d. Any of these  
19. Possible techniques for controlling filamentous organisms in an activated sludge process include:  
*a. Dosage of return sludge with a disinfectant such as chlorine or hypochlorite  
b. Lower DO levels in aeration bans so filamentous organisms cannot breathe or respire  
c. Lower F/M level to starve filamentous organisms  
d. Stop wasting to allow activated sludge bugs to gain control  
20. Nitrogen and phosphorous are biologically significant in waste water because:  
a. Nitrogen is reduced to pure nitrogen gas when phosphorous is present as a catalyst and this leads  
to nitrogen super saturation and rapid fish kills.  
b. Nitrogen in the form of nitrates will reduce sodium thiosulfate and thus cause low results in the  
D.O. test while phosphates will oxidize the iodine to iodide and cause high results.  
*c. Nitrogen and phosphorous are essential nutrients that can enable aquatic plant growth to be-  
come troublesome.  
d. Nitrogen pentaphosphate is toxic to the organisms that metabolize.  
21. What are the bacteria that remove ammonia in the activated sludge process  
a. Filamentous bacteria  
*b. Nitrosomonas  
c. Stalked ciliates  
d. E. Coli  
22. The loading on a nitrification system consists of ammonia plus organic nitrogen and is typically  
referred to as total nitrogen.  
a. True  
*b. False  
23. Chlorine is being applied at a constant dose rate of 24 mg/L to a partially nitrified activated sludge  
effluent having a pH of 6.8 and a temperature of 67 deg. F. Ammonia nitrogen is found to range  
from 2 to 3 mg/L in this effluent. Disinfection in this effluent might be difficult because:  
a. A temperature of 70 deg. F or higher is necessary in order to achieve effective disinfection.  
b. Chloramines are present most of the time.  
c. The chlorine dose rate is too low.  
d. The pH of this effluent will limit the effectiveness of free chlorine.  
*e. The ratio of chlorine to ammonia nitrogen may make it difficult at times to maintain adequate  
chlorine residual.  
24. During breakpoint chlorination:  
a. A ratio of 5 parts of chlorine to 1 part of ammonia nitrogen is adequate to reach the breakpoint.  
b. Ammonia nitrogen is reduced to nitrogen dioxide.  
*c. As the breakpoint is approached, additional chlorine causes the chlorine residual to decrease.  
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Chapter 12. Nutrient Removal  
d. Significant concentrations of dichloramine remain after the breakpoint is passed.  
25. In the breakpoint chlorination method of ammonia nitrogen removal, parts of chlorine are required  
for each part of ammonia removed.  
a. 3.0  
b. 6.7  
*c. 7.6  
d. 2.0  
12.3 Phosphorous removal  
207  
26. The correct statement regarding the effect of nitrogen in chlorine disinfection is:  
a. The reaction of ammonia with chlorine increases pH.  
b. The reaction of ammonia with chlorine decreases pH.  
c. The reaction of nitrite with free chlorine produces chloride.  
*d. The reaction of ammonia with free chlorine produces chloramines.  
e. The reaction of nitrate with free chlorine produces nitrogen gas.  
27. You are the superintendent of a 100,000 gpd conventional activated sludge plant which discharges  
into a shallow bay. Your NPDES permit currently sets a discharge limit of 5 mg/L for total ammonia-  
nitrogen in your effluent. However, the Regional Water Quality Control Board at the request of the  
Department of Fish and Game will soon revise your plant’s discharge requirements and have even  
lower total ammonia limits so that certain species of fish may be re-introduced into the bay. The  
proposed maximum total ammonia-nitrogen concentration limit for November through March 15th  
- 2.0 mg/L and 1.0 mg/L for the period March 16th through October 31st Answer the following  
questions:  
1. Limits on the concentration total NH3 - N are frequently placed on secondary effluents dis-  
charged into sensitive receiving waters. An important consideration in the setting of these  
limits by the RWQCB would be:  
a The concentration of algal cells in the receiving water.  
b The pH & temperature of the receiving water.  
c The salinity of the receiving water.  
d The population of nitrosomonas bacteria in the final effluent.  
e
The concentration of copper ions in the final effluent because these ions form a non-  
toxic complex with the ammonia nitrogen.  
2. Many NPDES permits limit the amount of ammonia nitrogen that can be discharged into a  
stream because:  
a Ammonia exerts an oxygen demand in the stream  
b Ammonia can be toxic to aquatic life  
c Ammonia reacts with chlorine which can interfere with disinfection  
d Any of these  
3. Impacts of new ammonia-nitrogen limits on plant operations would include:  
a Process change to remove ammonia further: nitrification – denitrification  
b Increased MCRT, higher activated sludge power demands and Lower sludge yields  
c Potential impacts to disinfection strategy related to breakpoint chlorination  
d Options [a] and [b]  
e Options [a], [b] and [c]  
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Chapter 12. Nutrient Removal  
28. You are the operations supervisor of a wastewater treatment plant that consists of the following  
unit processes: primary clarifiers, activated sludge aeration tanks, secondary clarifiers, a DAF  
thickener used to thicken WAS, and anaerobic digesters. Anaerobic digester gas is used, when it  
is available in sufficient amounts, to fuel engines that drive blowers supplying air to the activated  
sludge aeration tanks. Waste heat from these engines is sufficient to heat the digesters. When  
a sufficient amount of gas is not available, then electric motors are used to drive blowers. The  
regional water quality control board will soon impose a limit of 1.0 mg/L on the discharge of  
ammonia-nitrogen. In order to meet this limit, your plant will be modified to achieve complete  
nitrification.  
Answer the following questions:  
1. Demand Charge is measured in:  
a). $/kWh  
b). $/Hp  
c). $/kW  
d). $/day  
2. Most significant impact on energy consumption related to changes in mode of operation in  
order to meet the new permit requirement will be due to:  
a). Increased digester gas production  
b). Additional sampling and testing requirements  
c). Associated with meeting higher F:M requirements for nitrification  
d). Higher aeration air requirements  
3. Which of the following is true for Demand Charges:  
a). Applies to residential customers also.  
b). Based upon how much electricity used and the rate at which it is consumed  
c). It is calculated based on 15-minute interval data  
d). Options a) and c)  
e). Options b) and c)  
f). Options a), b) and c)  
4. Digester gas production is expected to:  
a). Increase.  
b). Decrease  
c). Remain the same  
5. More power will need to be purchased because of:  
a). Increased power demands related to denitrification  
b). Meet power demand associated with higher F:M ratio requirements  
c). Additional sludge pumping to digesters  
d). Additional RAS pumping requirement  
12.3 Phosphorous removal  
209  
29. Part A: You are the chief plant operator at a 3 MGD conventional activated sludge plant that dis-  
charges into a nearby creek. Currently you operate this plant at an MCRT of between 4 to 5 days.  
Your plant’s maximum daily total coliform limit is 23 MPN/100 ml. Chlorine disinfection to  
achieve a total chlorine residual of 5 mg/L followed by dechlorination with sulfur dioxide had  
been used to meet your effluent total cofifom, limit. Recently, at the request of the state department  
of fish and game, the RWQCB revised your plant’s NPDES pemiit in order to make the creek more  
habitable for certain species of fish. Your plant will now be required to meet a final effluent total  
ammonia-nitrogen (N-NH3) limit of 2 mg/L during the winter months (November 1- March 1) and  
1 mg/L during the rest of the year. Your plant’s total coliform limit was not revised.  
Answer the following questions:  
(a) Define total ammonia-nitrogen?  
(b) Why are such low limits necessary? Why are there different limits in winter and summer?  
(c) Identify and briefly discuss three significant impacts that these new limits will have on plant  
operations  
Part B: Prior to this revision of your NPDES permit, your plant rarely had a problem meeting its to-  
tal colifom, limit. Within days of meeting the revised ammonia-nitrogen limit, you’ve experienced  
frequent violations of your total coliform limit. Plant records show that maintaining a total chlorine  
residual of 5 mg/L has been difficult even though more chlorine has been used. Likewise, chlorine  
use has increased significantly. You’ve also noticed that at times the effluent in the chlorine contact  
chamber is "crystal clear" almost like a swimming pool. Plant records show no significant changes  
in plant flow or influent characteristics (i.e. BOD, TSS, pH, etc.).  
(a) How do these observations help explain why you are having total colifom, violations?  
(b) Identify and briefly discuss one step you might take to prevent these coliform violations.  
Response:  
Part A:  
a Define total ammonia-nitrogen  
Total ammonia nitrogen is the total amount of nitrogen in the forms of NH3 and NH+4 in  
the wastewater.  
b Why are such low limits necessary? Why are there different limits in winter and summer?  
Presence of nitrogen in wastewater effluent will promote plant & algae growth causing  
eutrophication (oxygen depletion) of the water body in which the effluent is discharge  
impacting aquatic life  
• Additionally, ammonia-nitrogen is toxic to aquatic life  
Toxicity of ammonia is affected by temperature, fish breeding times and susceptibility of  
juvenile fishes to ammonia– thus the different limits during winter and summer  
c
Identify and briefly discuss three significant impacts that these new limits will have on plant  
operations  
Process changes to remove ammonia further through implementation of nitrification –  
denitrification as part of the activated sludge process  
Potential impact to the disinfection process required to meet the coliform limit - break-  
point chlorination  
Associated impacts include:  
Increase MCRT – increased RAS pumping  
Lower sludge yields – less biosolids  
Additional oxygen requirements – more energy consumption – higher power costs  
210  
Chapter 12. Nutrient Removal  
Potential need for alkalinity and cBOD supplements for facilitating nitrifica-  
tion/denitrification  
Less digester gas production  
Part B:  
a How do these observations help explain why you are having total colifom, violations?  
• Potential impact of nitrite accumulation on chlorine disinfection  
• 5 ppm of chlorine used up for each part of nitrite present  
• Chlorine being consumed by nitrite and is not available for disinfection  
b Identify and briefly discuss one step you might take to prevent these coliform violations.  
• Measure nitrite levels at the effluent end of the activated sludge reactors  
• Increase DO for the nitrification step to facilitate the oxidation of nitrite to nitrate  
12.3 Phosphorous removal  
211  
30. Define nitrification. Name four factors that effect nitrification and discuss ranges, requirements and  
what lab tests are used.  
Response:  
Response:  
a Define nitrification  
In wastewater treatment, nitrification is used for the removal of ammonia nitrogen - the  
predominant form of nitrogen in wastewater. Nitrification involves a two step biological  
process where in the first step ammonia NH3 is biologically oxidized by ammonia  
oxidizing bacteria such as nitrosomonas to nitrite (NO2), followed by the oxidation of  
NO2 to nitrate (NO3) by ntirite oxidizing bacteria such as nitrobacter.  
b
Name four factors that effect nitrification and discuss ranges, requirements and what lab tests  
are used.  
i. Dissolved oxygen  
4.5 parts of O2 are needed for every part of NH4 +N (nBOD) to be degraded. In order for  
nitrification to occur, dissolved oxygen levels of 1.0 to 4.0 mg/L are usually maintained  
in the aeration tanks. Maximum nitrification occurs at DO levels of about 3.0 mg/L.  
Dissolved oxygen (DO)is measured using a DO probe.  
ii. Alkalinity and pH  
Presence of adequate alkalinity in the mixed liquor is critical as the nitrification pro-  
cess consumes alkalinity. Alkalinity provides a buffer for pH change. If the alkalinity  
is reduced beyond certain levels, further formation of acidic metabolic byproducts of  
bacterial activity may lead the pH to decline inhibiting bacterial activity. 7.14 parts of  
alkalinity are required for each part of ammonia removed. Nitrification rates are  
rapidly depressed as the pH is reduced below 7.0. pH levels of 7.5 to 8.5 are consid-  
ered optimal. An alkalinity of 60 mg/L in the secondary treatment reactor is generally  
required to ensure adequate buffering.  
Alkalinity is measured in the laboratory by titrating the sample with an acid until a  
specific pH is reached. Alkalinity is reported as mg/l of calcium carbonate.  
iii. MCRT, F/M, or Sludge Age  
For nitrification to occur the activated sludge treatment process needs to be operated  
at higher MCRT/sludge age (low F:M) as the reproductive rates of nitrfiers is low. An  
MCRT of greater than 8 days is typically considered essential for nitrification to occur.  
iv. Wastewater temperature  
Nitrification is inhibited at lower wastewater temperatures in wastewater treatment  
plants. To achieve the same level of nitrification, longer detention time may be needed  
in the winter versus the summer months since the activity drops significantly. Lower  
temperature effects on Nitrification may be partially mitigated by increasing MLVSS  
and MCRT. The optimal temperature range for nitrification is between 60to 95de-  
grees F. Below 40nitrification will probably not occur.  
v. Inhibition to nitrification by toxic compounds  
Many compounds can be toxic to nitrifiers- Nitrosomonas and Nitrobacter. These in-  
clude unionized ammonia, heavy metals, solvents and cyanide.  
212  
Chapter 12. Nutrient Removal  
31. What is meant by un-ionized ammonia and total ammonia nitrogen? Why is it important to set a  
low limit on un-ionized ammonia-nitrogen?  
Response:  
a What is meant by un-ionized ammonia and total ammonia nitrogen?  
Ammonia is the predominant form of nitrogen in wastewater. Ammonia (NH3) can exist  
as ammonia itself which is the unionized form or it could change to its ionized form - am-  
monium (NH+4 ) by absorbing a proton (H+). Total ammonia nitrogen is the sum of the con-  
centrations of the unionized ammonia and the ammonium ions present. These two forms of  
nitrogen can rapidly change from one to the other depending on pH and temperature.  
b Why is it important to set a low limit on un-ionized ammonia-nitrogen?  
i. Nitrogen in any of its forms, if present in wastewater effluent discharge, promote growth  
of plant and algal matter in the receiving waters causing destruction of the normal  
aquatic life mainly due to oxygen depletion - eutrophication. Ammonia is sought by  
nitrifying bacteria and is converted to nitrate at the expense of the oxygen present in the  
water body.  
ii. Additionally, ammonia specially in its unionized form is particularly toxic to aquatic life  
12.3 Phosphorous removal  
213  
32. Define and explain importance of:  
(a) Breakpoint chlorination  
(b) Chlorine demand  
(c) Chlorine residual  
(d) Chloromines  
(e) Rotometer  
Response:  
a Breakpoint chlorination  
When disinfecting with chlorine, as chlorine is added to wastewater, the presence of inorganic  
and organic substances including ammonia in the wastewater, will exert a demand for chlo-  
rine as chlorine is a strong oxidizing agent. This consumption of chlorine does not allow the  
chlorine to be present as free chlorine - the strongest form of chlorine disinfectant. Break-  
point chlorination is the point where the demand for chlorine has been fully satisfied and any  
further addition of chlorine will show a proportional increase in free chlorine residual.  
b Chlorine demand  
The amount of chlorine used up as part of the reaction of chlorine with the inorganic and  
organic substances present in wastewater is referred to as the chlorine demand.  
c Chlorine residual  
Chlorine residual is the sum of free chlorine and combined chlorine and it represents the  
amount of chlorine available for disinfection.  
d Chloramines  
Chloramines which include monochloramine, dichloramine and trichloramine are products of  
the reaction of chlorine with ammonia.  
e Rotometer  
Rotometer is a flow measurement device most commonly used for measuring the flow of  
chlorine gas for disinfection.  
13. Solids Treatment Overview  
13.1 Why do we need to treat wastewater solids?  
Sludge is generated from the wastewater treatment processes - settled solids and scum from pri-  
mary and secondary treatment processes  
• This sludge contain organic compounds and also elements that are beneficial plant nutrients  
• However, the organic solids in the sludge are not stable (i.e. they will decay) and include pathogens.  
Prior to disposal, sludge has to be treated – stabilized, so that its disposal or reuse does not pose a  
threat to public health.  
Sludge treatment is very critical as it is an expensive process and sludge disposal is subject to strict  
regulatory requirement.  
Even solids are only a small component of wastewater, the solids treatment and disposal account  
for a very substantial portion of wastewater treatment costs. Typically 40 to 60% of total wastewa-  
ter treatment operations cost is attributable to sludge treatment and disposal.  
NOTE: Solids removed during Preliminary Treatment, from barscreens and grit chambers are typi-  
cally not treated as part of the solids treatment process. These solids are disposed off at a landfill  
Typical solids treatment is comprised of the following three sequential steps:  
1. Sludge thickening  
2. Sludge stabilization  
3. Sludge dewatering  
13.2 Sludge thickening  
Sludge thickening involves the removal of excess water from the primary and secondary sludge increasing  
the solids content of the sludge and reducing the volume of sludge to be treated in the sludge stabilization  
 
 
 
216  
Chapter 13. Solids Treatment Overview  
process. Sludge thickening reduces the volume of sludge that need to be handled in the sludge stabiliza-  
tion step thereby reducing treatment cost.  
There is an upper limit of the solids concentration that can be effectively treated (stabilized) as  
increasing the solids concentration reduces its ability to be mixed and pumped easily. Typically the  
sludge thickening process produces sludge with a solids content of less than 10%.  
Benefits of thickening to the sludge stabilization process include:  
• Improved performance due to a lower volume of sludge  
• Cost savings in the construction of new facilities  
• Reduction in energy requirements as less water has to be heated  
Typical methods used for sludge thickening include:  
1. Gravity thickener - more suitable for primary sludge  
2. Dissolved air floatation thickener - more suitable for lighter, fluffier floc such as the secondary  
sludge.  
13.3 Sludge Stabilization  
Sludge stabilization process produces solids that are deemed safe for eventual disposal. Federal Part 503  
rule establishes requirements for the final use or disposal of sewage sludge. The solids disposal methods  
may include: land application, as a crop/vegetation fertilizer, placed on a surface disposal site for final  
disposal and fired in an incinerator.  
Biosolids is the term used for stabilized sludge which meets regulatory standards for beneficial reuse  
Sludge stabilization process results in the following:  
1. Reduction in amount of solids  
2. Pathogen reduction  
3. Odor reduction  
4. Reduction in vector attraction  
The main processes involved in sludge stabilization include:  
• Digestion - Aerobic or anaerobic  
• Lime or alkaline stabilization  
• Composting  
• Long term storage in lagoons  
• Thermal processes  
• Incineration  
13.4 Sludge Dewatering  
Solids stabilized using digestion process has only a small percentage by weight of solids -less than 5%.  
It therefore becomes necessary to dewater the stabilized sludge prior to hauling off-site for final disposal.  
Like thickening, the dewatering process does not treat the sludge. It increases the solids content to be-  
tween 15 to 30 percent and the higher solids content of the stabilized sludge makes it easier to handle and  
reduces costs associated with elements related to accomplishing the end objectives with the sludge – land  
application, composting, drying, incineration or landfill.  
Dewatering involves conditioning the sludge with a polymer and subjecting it to a physical process which  
include:  
1. Belt Filter Press  
2. Centrifuge  
 
 
14. Biosolids Regulations  
14.1 Biosolids Definition  
• Biosolids are treated solids produced as part of wastewater treatment  
• Biosolids can be disposed or recycled beneficially  
Biosolids must meet standards established under Title 40 of the Code of Federal Regulations  
(CFR), Part 503.  
• Grit and screenings are not considered as biosolids  
14.2 Biosolids Use/Disposal Methods  
1. Land application  
This is the most commonly used biosolids management method and is referred as a "recy-  
cling" option.  
This uses the organic and/or nutrient content of the biosolids to either condition and/or fertil-  
ize crops or other vegetation grown in the soil  
Land application allows for the beneficial utilization of soil-enhancing constituents such as  
plant nutrients and organic matter in the biosolids  
2. Surface disposal which requires availability of a large land which is lined with an impermeable  
material prior to the application of biosolids.  
3. Incineration where the biosolids is burnt to ash. This method utilizes its organic content to limit the  
amount of external fuel required to incinerate.  
14.3 Biosolids Regulations  
Part 503 rule applies to any person who applies biosolids to the land or fires biosolids in a biosolids  
incinerator, and to the owner/operator of a surface disposal site, or to any person who is a preparer  
or generator of biosolids for use, incineration, or disposal.  
 
 
 
 
218  
Chapter 14. Biosolids Regulations  
Table 5.1: Pollutant Concentration Limits and Loading Rates for Land Application  
• Part 503 standard includes:  
1. General requirements  
2. Pollutant limits  
3. Management practices  
4. Operational standards, and  
5. Requirements for the frequency of monitoring, record-keeping, and reporting  
14.4 Title 40 CFR Part 503 Requirements  
In order to land apply biosolids, each of the following three standards must be met.  
14.4.1 Pollutant concentration limits  
The biosolids produced must have concentrations of the 10 heavy metals listed, lower than the  
Ceiling Concentration thresholds.  
Sewage sludge exceeding the ceiling concentration limit for even one of the regulated pollutants is  
not classified as biosolids and, hence, cannot be land applied.  
Biosolids with heavy metal concentrations at or below the limit specified in the Pollution Concen-  
tration thresholds are classified as "High Quality Biosolids"  
Biosolids meeting pollutant concentration limits are subject to fewer requirements than biosolids  
meeting ceiling concentration limits  
14.4.2 Pathogen reduction  
• Pathogens are disease causing organisms such as bacteria, viruses and parasites  
 
 
 
14.4 Title 40 CFR Part 503 Requirements  
219  
The treatment method used for treating wastewater solids must meet standards related to pathogen  
reduction  
• Based upon the method used for treating wastewater solids, biosolids produced are classified as:  
Class A Biosolids  
The biosolids produced using the methods and standards identified are free of measurable pathogens.  
There are no pathogen related site restrictions for the application of Class A biosolids - including  
use in home gardens and landscaping  
Processes to further reduce pathogens (PFRP) treatment, such as those involving high temperature,  
high pH with alkaline addition, drying, and composting, or their equivalent are most commonly  
used to demonstrate that biosolids meet Class A requirements  
Class B Biosolids  
For Class B biosolids, pathogens have been reduced to levels that are unlikely to cause a threat to  
public health and the environment under specified use conditions.  
Processes to significantly reduce pathogens (PSRP), such as digestion, drying, heating, and high  
pH, or their equivalent are most commonly used to demonstrate that biosolids meet Class B require-  
ments  
Site restrictions are imposed on the application of Class B biosolids to ensure minimizing the  
potential for human and animal contact with the biosolids until environmental factors reduce the  
pathogens to below detectable levels.  
14.4.3 Vector attraction reduction  
Vector reduction standards are for preventing transmission of pathogens via rodents, birds, and  
insects from the land applied biosolids.  
• The vector reduction rule requirements are based on the following two approaches:  
1. Specifying organic matter decomposition processes viz., digestion and alkaline addition to  
mitigate vector attraction  
2. Biosolids sub-surface injection or incorporation within six hours so soil microbes out-  
competes/eliminates pathogens  
A minimum of 38% volatile solids reduction as part of the solids treatment process is a more common  
method of demonstrating compliance with vector attraction reduction requirements  
For biosolids to qualify for Exceptional Quality (EQ) Standards - the application of which is almost  
unrestricted, it must meet all three of the following:  
1. Pollutant concentration limits  
2. Class A requirements  
3. Vector attraction reduction standards  
 
220  
Chapter 14. Biosolids Regulations  
Chapter Assessment  
1. Grit and screenings from the preliminary treatment are also considered as biosolids  
*a. True  
b. False  
2. A true statement regarding the term “biosolids” is:  
a. The term is mandated for user by public law 92-500.  
b. The term was developed by US EPA to define all biologically toxic precipitates.  
*c. The term is recommended by WEF for “a primarily organic solids product, produced by  
wastewater treatment processes, that can be beneficially recycled”.  
d. The term is used by the California Water Resources Control Board to include “all insoluble  
matter derived from living aquatic organisms.  
3. When you spread sludge on agricultural land, the annual application rate of cadmium in the sludge  
should be less than 2 lbs/acre/year If your sludge contains 30 mg cadmium per kilogram of solids  
and your plant produces 950,000 lbs per year of dry solids, how many acres do you need?  
a. 3 acres  
*b. 14 acres  
c. 19 acres  
d. 27 acres  
222  
4. Essay type question:  
Chapter 14. Biosolids Regulations  
(a) Define vector and pathogens.  
(b) Class A & B landfill requirements: what you need to do to get your anaerobic digester to  
produce class B sludge to prepare for landfill.  
(c) What unit processes can produce class A & B sludge?  
Response:  
a Define vector and pathogens.  
• Pathogens are disease causing organisms such as bacteria, viruses and parasites  
Vectors are organisms such as rodents and insects that can carry disease by carrying and  
transferring pathogens  
b
Class A & B landfill requirements: what you need to do to get your anaerobic digester to  
produce class B sludge to prepare for landfill.  
For the digested sludge to qualify as Class B sludge it needs to meet the following related to  
40 CFR Part 503:  
• Meet the Minimum Concentration Standards of the Pollution Concentration Standards  
• Meet the digestion sludge detention time-temperature requirements  
• Meet the Vector Reduction Standards  
c What unit processes can produce class A & B sludge?  
For Class A sludge:  
i. Composting  
ii. Heat drying  
iii. Heat treatment  
iv. Thermophillic aerobic digestion  
v. Beta ray irradiation  
vi. Gamma ray radiation  
vii. Pasteurization  
14.4 Title 40 CFR Part 503 Requirements  
223  
For Class B sludge:  
i. Aerobic digestion  
ii. Air drying  
iii. Anaerobic digestion  
iv. Composting  
v. Lime stabilization  
15. Sludge Thickening  
Prior to the sludge stabilization process such as anaerobic digestion, the solids content of the sludge is  
increased by utilizing an appropriate thickening process.  
Notes:  
1) There is an upper limit of the solids concentration that can be effectively treated as increasing the solids  
concentration reduces its ability to be mixed and pumped easily. Digesters can effectively treat sludges  
with upto about maximum 7% - 9% solids.  
2) If a 5000 mg/l (0.5%) sludge is thickened to 5% solids concentration - it will reduce the volume of  
sludge by 90%  
15.1 Advantages of sludge thickening  
• Improved digester performance due to a lower volume of sludge  
• Capital Cost savings associated with less digester volume requirements  
• Operational costs savings - for sludge heating and mixing  
15.2 Sludge thickening methods  
1. Gravity thickener - more suitable for primary sludge  
2. Dissolved air floatation thickener - more suitable for lighter, fluffier floc such as the secondary  
sludge.  
15.3 Gravity thickener  
The gravity thickener is designed and operated similar to a circular primary and secondary clarifier.  
 
 
 
 
226  
Chapter 15. Sludge Thickening  
15.3.1 Principles of gravity thickener operations  
Upon entering the tank from the center, the sludge solids settle under the influence of gravity and  
these settled solids accumulate in the bottom of the gravity thickener as the sludge blanket. As  
the sludge becomes thicker it helps squeeze out more water from the sludge increasing the sludge  
solids content.  
Typical solids loading rates of a gravity thickener used for thickening primary sludge is 20-30 lbs  
TS/day-ft2  
 
15.4 Dissolved air floatation thickener (DAFT)  
227  
A picket fence-like mechanism which is attached to the bottom sludge rake arms is primarily to  
release entrapped gases. The sludge rake arms rotate at a very slow speed to ensure that it does  
not cause turbulenece causing the sludge to rise. The sludge is gently raked towards a sump in the  
center, from where the solids are withdrawn. The sludge rake encounter much higher torques than  
the typical primary clarifier rake, and is therefore designed to be more stronger and heftier.  
The thickened solids are drawn-off at regular intervals and the liquid fraction is decanted from the  
top and returned to the primary clarifier. The typical sludge concentration factor is about 2.  
• The sludge blanket is kept about three feet.  
The sludge blanket depth is typically maintained so that the sludge does not turn septic and the  
effluent is relatively clear and free of solids.  
Sludge Volume ratio (SVR) provides a means of regulating the detention time of sludge within the  
blanket of the thickener  
SVR is defined as the volume of the sludge blanket divided by the daily volume Of sludge pumped  
from the thickener.  
SVR is a relative measure of the average detention time of solids in the thickener and is calculated  
in days.  
• Typical SVRs for primary sludge range from 0.5 - 2 days.  
• Scum is removed from the top using a separate scum removal mechanism.  
• Chemical additives may be utilized to enhance settling and control odors.  
15.3.2 Elements of a gravity thickener  
• center feed column and baffle  
• drive assembly  
• scum removal system  
• thickened sludge pump  
• sludge rake with pickets  
• effluent weir  
• sludge hopper and pump  
15.3.3 Gravity thickener operational parameters  
• type and quality of sludge  
• hydraulic or surface loading rate (GPD/ ft2/day)  
• solids loading rate (lbs/ ft2/day)  
• sludge volume ratio - sludge detention time  
• solids and hydraulic loading rates, and  
• quantity and characteristics of the polymer used  
15.4 Dissolved air floatation thickener (DAFT)  
Opposite to the principle of the gravity thickener where the thickened sludge forms a blanket at the bot-  
tom, in the DAFT the thickened sludge forms a blanket on the top surface. Primary sludge is not generally  
suitable for thickening using DAFT as its solids are heavier and do not float easily. The WAS from the  
secondary clarifier which typically has a solids content of about 0.5% to 0.8% is thickened to about 4% to  
6% concentration – thickened WAS (TWAS) using the DAFT.  
 
 
 
228  
Chapter 15. Sludge Thickening  
15.4.1 Principles of DAFT operations  
WAS is conditioned with cationic polymer and introduced into the DAFT. DAFTs typically operate  
at a solids loading rate of 1-2 lbs TSS/hr-ft2. Polymer feed ranges from 5-15 lbs of polymer/ton of  
the WAS (feed) solids.  
Recycled water from the DAFT is pressurized with air in the saturation tank and mixed with poly-  
mer treated WAS as it is released at the bottom of the DAFT using the Back Pressure Control  
Valve.  
The dissolved air from the pressurized water is released as minute air bubbles rises upwards car-  
rying with it the polymer flocculated sludge to the surface. Adequate quantity of air is required to  
float the WAS solids. Air:Solids ratio is one of the key operating and control parameter. Typical  
air:solids ratios in a DAFT are between 0.03 - 0.05 lb air per lb TS. PS: Density of air used for  
air:solids ratio calculations - 0.075 lb air/ft3 air - this value is given as part of the problem.  
The thickened solids floating on the top are scrapped off the surface of the DAFT by flights into  
the TWAS sump from where it is pumped to the digesters. The subnatant - water below the solids,  
part of it is used for the air pressurization in the saturation tank and the remaining is the underflow,  
 
15.4 Dissolved air floatation thickener (DAFT)  
229  
which is returned back to the influent flow.  
The flight speed is critical to the DAFT performance. Fast flight speed would limit the thickness  
and density of the sludge blanket while slower flight speed would result in the thickened solids  
layer getting more dense and thick. Excessively thick solids layer could result in the solids escap-  
ing through the underflow.  
15.4.2 Elements of a DAFT  
• pressurization or saturation tank  
• thickened sludge skimmer with drive assembly  
• polymer dosing and injection system  
• thickened sludge pump  
• back pressure control valve  
• underflow removal  
• recycled flow system  
15.4.3 DAFT operational parameters  
• saturation pressure  
• solids loading rate (lbs TSS/hr- ft2)  
• hydraulic loading rate (GPD/ ft2/day)  
• feed solids concentration  
• detention period  
• air-to-solids ratio (lb air: lb solids)  
• type and quality of sludge  
• flight speed  
• solids and hydraulic loading rates, and  
• quantity and characteristics of the polymer used (lbs polymer/dry ton solids)  
 
 
Chapter Assessment  
1. In a gravity thickener the depth of the sludge is kept minimal (<six inches) to avoid solids going  
over the effluent weir  
a. True  
*b. False  
2. Sludge thickening is primarily conducted to reduce costs associated with biosolids hauling  
a. True  
*b. False  
3. Gravity thickener is commonly used for sludge dewatering  
a. True  
*b. False  
4. Sludge thickening is primarily conducted to reduce costs associated with biosolids hauling  
a. True  
*b. False  
5. A DAF thickener has effluent solids of 55 mg/L and float solids of 2.0%. Solids loading and poly-  
mer dosing is in the normal range. This data likely indicates:  
a. This unit is operating normally  
b. Too low air to solids ratio  
c. Float blanket too thick  
*d. Flight speed too fast  
e. Flight speed too slow  
6. An air flotation thickener will produce a thin float if:  
*a. Flight speed too high and skimmer wiper not adjusted properly  
b. Excessive air/solids ratio and polymer dosages too low  
c. High dissolved oxygen and flight speed too low  
d. Polymer dosages too high and unit overloaded  
7. An increase in the pool depth of a scroll-type centrifuge:  
232  
Chapter 15. Sludge Thickening  
a. would not affect the moisture content of the cake.  
b. would produce a drier cake  
*c. would produce a wetter cake, but produce a greater solids recovery.  
d. would not affect either solids recovery, nor cake moisture content.  
e. would require an increase in the cationic polymer dosage.  
8. A sludge thickened from 1% to 4% solids will be reduced in volume by how much?  
a. no more than 4% of original volume  
b. approximately 17% of original volume  
*c. approximately 25% of original volume*  
d. more information is needed  
9. Gravity thickeners, compared to DAFs, are best suited to:  
*a. Thickening primary sludge.  
b. Thickening waste activated sludge.  
c. Controlling sulfide odors.  
d. Removing filamentous bacteria.  
e. Provide highest concentration sludge.  
10. Which of the following is not the main reason for thickening sludge  
a. Improved digester performance due to a lower volume of sludge  
b. Cost savings in the construction of new digestion facilities  
c. Reduction in anaerobic digestion heating requirements since less water has to be heated  
*d. Reduce costs of biosolids hauling  
11. What zone is not involved in a belt filter press?  
a. Gravity  
b. Low pressure (wedge)  
c. High pressure  
*d. Twilight  
12. The float blanket in a DAF unit appears well flocculated and concentrated. Too low a flight speed  
would likely result in:  
a. Using excessive amounts of air.  
b. Float solids that are too thick.  
c. Too low an air-to-solids ratio.  
*d. Poor thickener underflow quality.  
e. De-flocculation of float solids.  
13. The least critical operational control of a dissolved air flotation thickener in producing an ade-  
quately thickened sludge is the:  
a. Flight speed.  
b. Air to solids ratio.  
c. Polymer dosage.  
*d. Recycle ratio.  
e. Pre-thickened WAS concentration.  
14. The operational control of a dissolved air flotation thickener most critical for producing an ade-  
quately thickened float solids is the:  
a. Fight speed.  
*b. Air to solids ratio.  
c. Polymer dosage.  
15.4 Dissolved air floatation thickener (DAFT)  
233  
d. Recycle ratio.  
e. Concentration of WAS being thickened.  
15. To increase solids recovery on a dual belt, belt press, the operator should:  
a. Increase the differential belt speed.  
b. Increase belt speed.  
c. Increase sludge feed rate.  
*d. Decrease sludge feed rate.  
e. Increase polymer feed.  
16. Too high a flight speed in a DAF will likely result in:  
a. Using excessive amounts of air.  
b. Excessive underflow volume.  
*c. Thin float solids.  
d. Too high an air to solids ratio.  
e. Too low an air to solids ratio.  
17. Which one of the following statements is TRUE in regard to DAF thickeners?  
*a. The air to solids ratio in a DAF unit is typically 0.03 to 0.05.  
b. The speed of the so-called "flights" has very little effect on the concentration of the float solids.  
c. Adjustments in the air to solids ratio in a DAF unit will affect the float solids but not the unit’s  
effluent suspended solids.  
d. Anionic polymers are typically used to condition the WAS feed to a DAF unit.  
18. Which one of the following process units is usually classified as a sludge thickening device as  
opposed to a dewatering device:  
*a. DAF unit.  
b. Sludge drying bed.  
c. Vacuum filter press.  
d. Belt press.  
e. All of the above are thickeners, not dewatering devices.  
19. Which one the following statements is TRUE in regard to gravity thickeners?  
a. Longer solids detention times are desired during summer operation of these units.  
b. The sludge-volume-ratio (SVR) or sludge detention time is defined as the volume of the sludge  
blanket divided by the daily volume of sludge withdrawn from the thickener.  
c. SVRs should be in range of 5 to 10 hours.  
*d. A likely cause of a gravity thickener producing a poor quality effluent is too low of a sludge  
blanket.*  
20. A DAF thickener has effluent solids of 55 mg/L and float solids of 2.0%. Solids loading and poly-  
mer dosing is in the normal range. This data likely indicates:  
a. This unit is operating normally  
b. Too low air to solids ratio  
c. Float blanket too thick  
*d. Flight speed too fast  
e. Flight speed too slow  
21. An increase in the pool depth of a scroll-type centrifuge:  
a. would not affect the moisture content of the cake.  
b. would produce a drier cake  
*c. would produce a wetter cake, but produce a greater solids recovery.  
234  
Chapter 15. Sludge Thickening  
d. would not affect either solids recovery, nor cake moisture content.  
e. would require an increase in the cationic polymer dosage.  
22. A sludge thickened from 1% to 4% solids will be reduced in volume by how much?  
a. no more than 4% of original volume  
b. approximately 17% of original volume  
*c. approximately 25% of original volume*  
d. more information is needed  
23. Gravity thickeners, compared to DAFTs, are best suited to:  
*a. Thickening primary sludge  
b. Thickening waste activated sludge  
c. Controlling sulfide odors  
d. Removing filamentous bacteria  
e. Provide highest concentration sludge  
24. Identify the incorrect statement regarding gravity thickeners.  
a. Gravity thickeners are similar in design to primary clarifiers.  
*b. The sludge blanket depth and the rate of sludge withdrawal are used to calculate the sludge  
detention time.  
c. The purpose of pickets in a sludge thickener is to gently stir settling sludge particles to release  
gases that may prevent the sludge particles from compacting.  
d. Mixtures of primary and waste activated sludge are never thickened in a gravity thickener due to  
the possibilility of de-nitrification.  
e. Solids loading (pounds per day per square foot) is an important guideline for a gravity thickener.  
25. On a routine check of a DAFT unit the operator finds suspended solids of 450 mg/L in the effluent.  
The float blanket appears well flocculated and concentrated. The operator should:  
a. increase the flight speed  
b. do nothing, the unit is operating normally  
c. reduce the air to solids ratio  
*d. increase the air to solids ratio  
e. check unit operating pressure  
26. Gravity-thickened primary sludge will contain solids within which of the following concentration  
ranges?  
a. 1,000 - 8,000 mg/l  
b. 10,000 – 40,000 mg/l  
c. 40,000 - 80,000 mg/l  
d. 100,000 - 400,000 mg/l  
*e. none of the above  
16. Solids Stabilization - Digestion  
• Sludge digestion is a microbiological process to treat - stabilize, the wastewater solids  
As part of digestion, the organic material present is biodegraded by microorganisms. It is the most  
common sludge stabilization method.  
There are two major sludge digestion processes - aerobic digestion which utilizes aerobic microor-  
ganisms and anaerobic digestion which utilizes anaerobic microorganisms.  
Comparison of aerobic and anaerobic digestion:  
Methane production (fuel use potential)  
Energy consumption  
Unlike anaerobic digestion, aerobic digestion does not produce methane  
Aerobic digestion has higher energy needs as air supply is required to  
support the aerobic activity. The energy needs for an anaerobic digester is  
limited to mixing the digester and maintaining digester temperatures  
Anaerobic digestion produces only about 20% of the biomass produced by  
Biomass production  
aerobic digestion  
Dewaterability of sludge produced  
Compared to anaerobic digestion, aerobic digestion produces sludge  
with poor dewaterability which leads to producing cake with lower solids  
content and therefore more expensive to haul  
Siginifcant ammonia is produced as part of anaerobic digestion which  
Ammonia concentration  
imposes additional ammonia removal needs particularly for plants with  
nitrogen discharge thresholds  
Sludge odors  
Capital cost  
Aerobic digestion has much less odor issues compared to anerobic diges-  
tion  
Initial capital costs are higher for anaerobic digesters due to higher SRT  
requirements is higher - methanogens are slower growing  
Process efficiency of aerobic digester drops in cold weather. The anerobic  
Weather impacts  
digestion is not affected by weather as the sludge in the digester is main-  
tained at the desired temperature  
For this class we will focus only on anaerboic digestion - the most common method - as anaerboic diges-  
tion generates digester gas which can be used as fuel and also due to its lower energy demands.  
 
236  
Chapter 16. Solids Stabilization - Digestion  
Figure 16.1: Floating Dome Anaerobic Digestion  
16.1 Digester design  
• The anaerobic digester is typically a large cylindrical concrete tank  
• The digester is operated as a continuous process at a fixed volume  
As sludge is fed into the digester it displaces an equal amount of sludge which leaves the digester  
through the digester overflow system - see the digester overflow system design diagram below  
The digester overflow system is designed to allow for selecting the level from which the sludge  
overflows  
The sludge typically occupies 70 - 90% of the total digester volume and the methane carbon diox-  
ide gas mixture occupies the headspace from where it is withdrawn also on a continuous basis.  
• The digester can be constructed with either a fixed or floating roof.  
16.2 Types of anaerobic digestion systems  
Low-Rate  
No mixing provided  
Long detention time of 30 to 60 days  
Suitable for low organic loading  
Intermittent sludge feed  
Stabilized solids settle to the bottom and are re-  
moved periodically  
Suitable for small wastewater treatment plants  
 
 
16.2 Types of anaerobic digestion systems  
237  
Figure 16.2: Fixed Dome Anaerobic Digestion  
Figure 16.3: Digestion Overflow System  
238  
Chapter 16. Solids Stabilization - Digestion  
High-Rate  
Mixing and external heating provided  
Uniform feeding  
Typical detention times of 10 to 20 days  
Most common  
Two-stage  
Digestion occurs in the heated and  
mixed primary digester vessel  
The unheated secondary tank  
serves as storage for digested solids  
and as a source of seed  
Summary Conditions for Anaerobic Sludge Digestion  
Temperature and detention  
Psychrophillic  
50 - 65 F 50 - 180 days  
time  
Mesophillic - Normal range  
Mesophilic Most common  
68 - 113 F  
95 - 98 F 15 - 30 days  
Thermophillic  
Optimal  
113 - 135 F 5 - 12 days  
7.0 to 7.1  
pH  
General Range  
Optimal  
6.4 to 7.4  
VS loading rate (high rate  
0.1 and 0.2 lbs VS/day/ ft3  
digester)  
Gas production  
Per pound volatile solids added  
8-12 cu. ft  
Per pound volatile solids destroyed  
Methane  
16-18 cu. ft.  
65%  
Gas Composition  
Carbon dioxide  
35%  
Hydrogen sulfide  
Normal  
Trace  
200-800 mg/L  
Volatile acid concentration  
(as acetic acid)  
Alkalinity Concentration  
Maximum  
Normal  
2000 mg/L  
2,000-3,500 mg/L (as calcium carbonate)  
3,000-5,000 mg/L (as bicarbonate)  
Volatile acids to alkalinity  
ratio  
Desirable  
< 0.25  
Upset condition  
> 0.4  
16.3 Digestion parameters and control  
239  
16.3 Digestion parameters and control  
16.3.1 Digestion temperature  
The anaerobic digestion is temperature sensitive and temperature of the sludge in the digester needs  
to be controlled to ensure proper operation.  
The activity and type of bacteria present in the digester is dictated by the operating temperature of  
the digester.  
Anaerobic digestion can be in the following three temperature ranges, each of which has its own  
unique microbiology.  
i. PSYCHROPHILIC DIGESTION  
If operating in this temperature regime, the digesters are maintained between 50 - 65 F. This  
regimen is not commonly used. Digestion in this range requires 50 to 180 days depending on  
the target volatile matter reduction  
ii. MESOPHILIC DIGESTION:  
These digesters operate between 68 - 113 F. Most common operating temperatures for  
mesophilic digesters is between 95 – 98 F and the typical number of days required for di-  
gestion is between 15 to 30 days.  
iii. THERMOHILLIC DIGESTION:  
These digesters’ optimal operating temperatures range is between 113 - 135 F and it typically  
requires 5 to 12 days.  
A digester does not tolerate temperature fluctuations and need to be maintained at stable tempera-  
tures for a significant duration.  
• As a rule of thumb temperature of a digester should not be changed more than 1 per day.  
• Temperature fluctuations will destroy the microbiological balance leading to digester failure.  
• External heating is provided to maintain sludge temperatures in the digester.  
• Temperature fluctuations also can be minimized by feeding the system at frequent intervals.  
An external or internal heat exchanger or direct steam injection is typically used for maintaining the  
sludge temperature  
16.3.2 Digestion mixing  
A sludge mixing system is integral to the digester design. It performs three main functions:  
1. It allows for the bacteria to properly come in contact with the food.  
2. It prevents accumulation of grit and scum in the digester which could lead to dead zones.  
3. It allows for maintaining uniform temperature. Typical mixing system designs achieve a turnover  
time of 30 to 45 minutes.  
16.3.3 Digestion feed  
• The sludge feed to the digester has a total solids content of about 3 – 6%.  
• 70% of the total solids fed are organic solids.  
• These organic solids are measured as volatile solids (VS)  
Hydraulic or Sludge retention time (SRT) is calculated by dividing the digester volume by the daily  
sludge flow  
Digester Volume (ft3 or gallons)  
Digester SRT (days) =  
Sludge Feed ( ft3 or gallons per day)  
 
 
 
 
240  
Chapter 16. Solids Stabilization - Digestion  
• The SRT in an anaerobic digester typically ranges from 10 to 20 days.  
A minimum SRT is essential to the digestion process to ensure that the necessary microorganisms  
are being produced at the same rate as they are removed from the system each day  
Accumulation of grit in the bottom or scum at the top of the digester would effectively reduce the  
working volume of the digester  
The amount of digester influent being pumped and the percent VS of the waste are a measure of the  
digester’s organic loading rate (influent mass per time)  
Control of the amount of volatile solids loading is one of the key critical operational control param-  
eter.  
Volatile solids loading rate of the digester is the mass of VS added to the digester each day divided  
by the operating volume of the digester – lbs VS/day/ft3  
• Typical high rate digesters VS loading range between 0.1 and 0.2 lbs VS/day/ ft3  
• The rate of digestion depends on the type of organic material present.  
The constituents of the organic material in primary sludge is typically feces, and plant and animal  
origin material. This organic material is relatively easy to digest compared to the organics in the  
secondary sludge which is primarily living and dead microorganisms and metabolic byproducts  
associated with microbiological growth and decay.  
Typical primary sludge from a treatment plant with preliminary treatment contains approximately  
65% organic material. 80% of this organic material is typically destroyed in the digester. Whereas,  
the secondary sludge typically contains 90% organics and only 60% of the secondary sludge organ-  
ics is destroyed.  
16.3.4 Volatile solids breakdown  
The volatile solids content of the sludge entering and leaving the digester are measured to quantify  
the solids removal in the digester  
The expected reduction of volatile solids of a properly operating digester is 40-60% of the total  
volatile solids present in raw sludge feed.  
The volatile solids in the feed sludge would be about 70% to 75% while the digested sludge would  
be 45% to 50% volatile solids  
The expected reduction of volatile solids of a properly operating digester is 40-60% of the total  
volatile solids present in raw sludge feed  
• The volatile solids reduction of the digester is provided by the Van Kleeck equation  
VSin VSout  
Digester VS reduction(%) =  
100  
VSin VSin VSout  
Digester volatile solids concentration is typically expressed as a percentage of the sludge total  
solids  
• 70% VS which means that 70% of the total solids is volatile solids  
The value of VSin and VSout for the digester VS reduction (Van Kleek) equation above should be in  
fraction and not as a percentage.  
For example: for a 70% VS content use 0.7 in the equation. Likewise 0.525 is used for 52.5% VS  
concentration  
• Higher volatile solids reduction implies higher gas production and lower biosolids hauling costs  
Breakdown of volatile matter in the sludge ultimately into methane (CH4) and carbon dioxide  
(CO2) occurs in multiple steps involving different groups of microorganisms.  
 
16.3 Digestion parameters and control  
241  
Step 1. This is the hydrolysis step where the complex organic matter in the sludge including carbo-  
hydrates, proteins, lignin, and lipids are converted to simpler compounds including sugars, soluble  
fatty acids and amines.  
Step 2. This involves formation of volatile acids from the products from Step 1. This is the acid  
formation step.  
Step 3. The acid formed in Step 2 is converted into methane and carbon dioxide. This step is ac-  
complished by a group of methane forming bacteria.  
The digester conditions are maintained to ensure optimal conditions conducive to the activity of the  
methane formers.Good digester operations requires ensuring that conditions are kept favorable for  
the methane formers  
16.3.5 Digester pH and alkalinity  
The anaerobic digestion requires a symbiotic relationship between the different groups of microor-  
ganisms involved.  
Each group of organisms involved in the anaerobic digestion process have an optimal pH for maxi-  
mum rate of reaction –  
Hydrolysis: pH 5-7 optimal  
Acid formation: pH 5-7 optimal –  
Methane formation: pH 7-8 optimal, pH 6.5-8.5 operational  
In a healthy digester the volatile acids produced in Step 2 are used in Step 3 as food by the methane  
formers at about the same rate as they are produced.  
As the methane forming microorganisms in the digesters are very sensitive to the digester con-  
ditions including organic content, pH, temperature and toxins and also grow very slowly, it is  
necessary to maintain the digester pH close to neutral  
The conversion of the volatile fatty acids to methane by methane formers allow for controlling the  
accumulation of volatile acids  
If the activity of methane formers is suppressed due to conditions such as low pH, volatile acids  
will accumulate potentially causing the digester to become "sour"  
In a normally operating digester, the alkalinity present in the sludge prevents the pH from dropping  
due to the formation of volatile acids during Step 2.  
The alkalinity consumed due to the acids formed in Step 2 is replenished in Step 3 by the formation  
of bicarbonate from the CO2 and ammonia produced during digestion.  
NH3 +H2O+CO2 NH4HCO3  
Unlike the acid formers, Thus, if the digester condition such as the pH was to change – drop be-  
cause of the formation of acids and a lack of alkalinity, Step 3 will not happen resulting in the  
digester pH dropping even further resulting in a “sour” digester condition.  
• When acid formers out produce the methane formers the volatile acids increase sharply  
The methane formers are the key to digester operation and are very vulnerable to low pH There  
needs to be sufficient alkalinity present to prevent pH changes as the volatile fatty acids are being  
formed.  
If sufficient alkalinity is not present the pH will drop further inhibiting the activity of methane  
formers and the digester will turn “sour”  
The bicarbonate ion (HCO3 ) is the main source of buffering capacity to maintain the system’s pH  
 
242  
Chapter 16. Solids Stabilization - Digestion  
in the range of 6.5 – 7.6. The concentration of HCO3 in solution is related to the percent of carbon  
dioxide in the gas phase  
Alkalinity consumed by the acid production is offset by alkalinity produced as part of the methanogenic  
activity  
Regular monitoring the digester alkalinity and volatile acids is vital in ensuring a stable digestion  
process.  
In a well operating digester the volatile acids expressed as acetic acid would be in the range of 50  
to 500 mg/L and the alkalinity expressed as calcium carbonate would be in the range of 2,000 to  
3,000 mg/L (bicarbonate alkalinity of 2,500mg/L and 5,000mg/L)  
The volatile acid-to-alkalinity ratio indicates if the digester has enough buffering capacity for the  
volatile acids being produced.  
Increase in volatile acid to-alkalinity ratio will increase the potential for pH decrease which would  
result in an upset digester.  
In general keeping the volatile acids to alkalinity ratio at 0.25 or less is desirable for good opera-  
tions.  
• Ratios above 0.4 indicate upset and the need for corrective action.  
16.3.6 Digester pH control  
Following chemicals: calcium oxide, calcium hydroxide, anhydrous ammonia, Ammonia (liquid),  
ammonium hydroxide, sodium carbonate, sodium bicarbonate, and sodium hydroxide have been  
used for increasing the pH of a sour digester.  
The use of calcium oxide (quicklime) or calcium hydroxide (slaked or hydrated lime) is probably  
the most dangerous if mixed with water prior to feeding. If instead of adding these two chemicals  
to water for digester feeding, adding water to these chemicals would pose the threat of a violent  
reaction and splattering of this strong caustic.  
Excess lime should be avoided as lime reacts with carbon dioxide to form calcium carbonate.  
Ca(OH)2 +CO2 CaCO3 +H2O  
If carbon dioxide is removed too rapidly or in too large a quantity from the sludge, then carbon  
dioxide from the biogas will replace the carbon dioxide lost from the sludge. When carbon dioxide  
is lost from the biogas, a partial vacuum condition develops under the digester dome. This con-  
dition may cause the digester cover to collapse. Also, as the concentration of alkalinity increases  
in the anaerobic digester, the continued use of quick lime results in the precipitation of calcium  
carbonate.  
• Sodium hydroxide would be the next most dangerous followed by ammonium hydroxide.  
The use of anhydrous ammonia should be limited to facilities that have the equipment to handle gas  
cylinders and that have feed connections or the ability to make the necessary feed connections.  
The feeding of lime can only raise the digester pH to about 6.8. When lime is added it reacts with  
carbon dioxide to form calcium bicarbonate. Excessive feeding of lime causes insoluble calcium  
carbonate to form. In addition excessive lime may remove too much carbon dioxide which could  
lower gas pressures or even the formation of a vacuum. This could cause air to be drawn into the  
digester and cause an explosive gas mixture.  
16.3.7 Digester gas production  
In the anaerobic digestion process microorganisms convert volatile matter into mainly methane  
(CH4) and carbon dioxide (CO2)  
 
 
16.4 Digester safety  
243  
Typical composition of digester gas is about 58 - 65% methane and 30 - 35% CO2 it also includes  
traces of ammonia nitrogen, hydrogen sulfide, and other gases.  
• Monitoring digester gas production is one of the more important operational parameter.  
Gas production ranges between 10 to 16 cubic feet per pound of volatile matter destroyed and the  
gas production remains stable over time.  
Low gas production is an indicator of digester issue related to - toxicity, temperature, volatile acid  
to alkalinity ratio, mixing, or feed rates.  
16.3.8 Digester failure  
• Digester failure conditions are marked by conditions which include increases in:  
volatile acids  
volatile acids: alkalinity ratio  
drop in pH  
drop in gas production, and  
an increase in CO2 concentration  
The failure is typically attributable to factors including overloading, toxicity and temperature  
control issues.  
• Digester failure can be controlled by:  
stopping or reducing the sludge feed  
exercising pH controls by adding alkaline chemicals, and  
through reseeding the digester with sludge from another healthy digester  
16.3.9 Digester toxicity issues  
• Toxicity could severely impact a digester performance.  
Contributors to toxicity issues include high concentrations of ammonia and toxic metals such as  
arsenic, cadmium, chromium (hexavalent), copper, nickel, zinc.  
Ammonia toxicity: Ammonia from industrial sources, organic overloading, or from using anhy-  
drous ammonia to correct a digester pH problem can cause a die-off of bacteria; ammonia toxicity  
begins around 1,500 mg/L and is totally toxic at 3,000 mg/L.  
16.3.10 Nutrients and trace constituents requirement  
The major nutrients required for anaerobic digestion are nitrogen and phosphorus. An average cell con-  
tains approximately 12.5 percent nitrogen and 2 percent phosphorus. Sodium, potassium, calcium, magne-  
sium, chloride, and sulfate ions are also required for proper digestion.  
16.4 Digester safety  
A. Pressure & Vacuum Protection:  
In a digester,as the gas production and sludge withdrawal is continuous, there exists a poten-  
tial to create pressurized or vacuum conditions within the digester which could jeopardize the  
structural integrity of the digester.  
Typically the gas in the digester dome is maintained at a pressure of 8-9 inches of water  
column.  
A pressure relief valve is designed into the digester dome which allows to protect the digester  
from excessive pressure or vacuum conditions.  
 
 
 
 
244  
Chapter 16. Solids Stabilization - Digestion  
In the event of the digester dome being under excessive pressure, the valve will open to  
relieve the pressure and likewise if the valve senses a low pressure or vacuum condition, it  
will open up just enough to let in the air to bring the gas in the digester dome back to normal,  
set pressure.  
B. Fire and Explosion Protection:  
• For a fire or explosion to occur, following three conditions must be met simultaneously:  
(a) Presence of fuel – combustible material  
(b) Oxygen (air) must exist in certain proportions, and  
(c) Ignition source, such as a spark or flame.  
• Digester gas contains methane which is a combustible gas.  
The ratio of fuel and oxygen that is required varies with each combustible gas or vapor. The  
Lower Explosive Limit (LEL) and (UEL) for methane is 5% and 15% respectively.  
A 5% LEL implies that a methane concentration in air of less than 5% is too “lean” to burn  
and a 15% UEL implies that a concentration of methane exceeding 15% in air would be too  
“rich” to burn. So for methane, the flammable range is between 5% and 15%. As the digester  
gas methane content is typically between 60 to 65%, dilution of digester gas with air has the  
potential to form an explosive mixture.  
A flame arrestor is typically installed at the outlet of the digester gas piping to protect the  
digester from any external heat or ignition source. The flame arrestor is essentially a heat  
exchanger which cools down the ignition source or flame entering the digester.  
16.4 Digester safety  
245  
Digester Pressure Relief  
Digester Flame Arrestor  
Chapter Assessment  
1. A change in the-pH-of digesting sludge in anaerobic digester is the best early warning indicator of  
potential digester upset.  
a. True  
*b. False  
2. A good maintenance program should be established for all flame arresters to ensure they are all set  
at the recommended "pop-off’ pressures.  
*a. True  
b. False  
3. A healthy anaerobic digester should have a carbon dioxide concentration of more than 40%.  
a. True  
*b. False  
4. A high rate anaerobic digester is always heated and mixed.  
*a. True b.False  
5. Anaerobically digested sludge produces gas as a by-product. The gas produced is of little or no  
value.  
a. True  
*b. False  
6. Digester gas containing 60% methane by volume will likely explode when exposed to a spark or  
flame.  
a. True  
*b. False  
7. Digester gas containing 60% methane by volume will likely explode when exposed to a spark or  
flame  
*a. True  
b. False  
8. In an anaerobic digester, the [1] is destroyed producing [2]. The digester gas production is typically  
248  
Chapter 16. Solids Stabilization - Digestion  
in the range of [3] cubic feet of digester gas per lb of volatile matter destroyed.  
9. Gas production in an anaerobic digester results from the destruction of fixed solids  
a. True  
*b. False  
10. A well operating digester will have a CO2 concentration of greater than 60%  
a. True  
*b. False  
11. Volatile solids removal efficiency of a digester is most commonly monitored utilizing the BOD test  
a. True  
*b. False  
12. Volatile solids removal efficiency of a digester is most commonly monitored utilizing the BOD test  
a. True  
*b. False  
13. The range of volatile material in raw sludge may be as high as 60% - 80% while digested sludge  
normally will be below 20%.  
a. True  
*b. False  
14. Foaming in digesters is often due to rapid acid digestion is caused by adding too much raw sludge  
during too short a time period.  
*a. True  
b. False  
15. Aerobic digestion produces methane gas that provides energy for other operations.  
a. True  
*b. False  
16. A good maintenance program should be established for all flame arresters to ensure they are all set  
at the recommended "pop-off" pressures.  
a. True  
*b. False  
17. A flame arrester in the gas line between a waste gas burner and an anaerobic digester contains  
plates which should periodically be inspected and cleaned.  
*a. True  
b. False  
18. The methane forming bacteria in an anaerobic digester reproduce more rapidly than the acid form-  
ing bacteria.  
a. True  
*b. False  
19. The main function of a secondary digester in normal operation is to provide a storage space for  
seed sludge.  
a. True  
*b. False  
20. Supernatant liquor is withdrawn from the lowest possible level in an anaerobic digester.  
a. True  
*b. False  
21. The pH of digested sludge in a healthy anaerobic digester will be near 7.0  
*a. True  
16.4 Digester safety  
249  
b. False  
22. Volatile solids reduction is a measure of the effectiveness of an anaerobic digester.  
*a. True  
b. False  
23. A properly operated anaerobic, digester functions best within a pH range of 5.4 to 5.8.  
a. True  
*b. False  
24. In computing anaerobic digester loadings, it is necessary to take into account the solids lost in the  
supernatant system.  
a. True  
*b. False  
25. In conventional secondary wastewater treatment processes, aerobic decomposition of solids will  
occur.  
*a. True  
b. False  
26. Aerobic digesters are operated under the principle of extended aeration from the activated sludge  
process.  
*a. True  
b. False  
27. After pumping raw sludge to a digester, it is necessary to seal both ends of the line to prevent back  
siphonage.  
a. True  
*b. False  
28. Propeller mixers are located mainly on floating covers of anaerobic digesters rather than on fixed  
covers.  
a. True  
*b. False  
29. A gallon of water weighs 8.34 pounds. A gallon of digested sludge, due to its high volatile content,  
will weigh much less.  
a. True  
*b. False  
30. The range of volatile material in raw sludge may be as high as 60% - 80% while digested sludge  
normally will be below 20%.  
a. True  
*b. False  
31. Foaming in digesters is often due to rapid acid digestion is caused by adding too much raw sludge  
during too short a time period.  
*a. True  
b. False  
32. Aerobic digestion produces methane gas that provides energy for other operations.  
a. True  
*b. False  
33. A good maintenance program should be established for all flame arresters to ensure they are all set  
at the recommended "pop-off" pressures.  
a. True  
250  
Chapter 16. Solids Stabilization - Digestion  
*b. False  
34. A flame arrester in the gas line between a waste gas burner and an anaerobic digester contains  
plates which should periodically be inspected and cleaned.  
*a. True  
b. False  
35. The methane forming bacteria in an anaerobic digester reproduce more rapidly than the acid form-  
ing bacteria.  
a. True  
*b. False  
36. The main function of a secondary digester in normal operation is to provide a storage space for  
seed sludge.  
a. True  
*b. False  
37. Supernatant liquor is withdrawn from the lowest possible level in an anaerobic digester.  
a. True  
*b. False  
38. The pH of digested sludge in a healthy anaerobic digester will be near 7.0  
*a. True  
b. False  
39. Volatile solids reduction is a measure of the effectiveness of an anaerobic digester.  
*a. True  
b. False  
40. A properly operated anaerobic, digester functions best within a pH range of 5.4 to 5.8.  
a. True  
*b. False  
41. In computing anaerobic digester loadings, it is necessary to take into account the solids lost in the  
supernatant system.  
a. True  
*b. False  
42. In conventional secondary wastewater treatment processes, aerobic decomposition of solids will  
occur.  
*a. True  
b. False  
43. Aerobic digesters are operated under the principle of extended aeration from the activated sludge  
process.  
*a. True  
b. False  
44. After pumping raw sludge to a digester, it is necessary to seal both ends of the line to prevent back  
siphonage.  
a. True  
*b. False  
45. A gallon of water weighs 8.34 pounds. A gallon of digested sludge, due to its high volatile content,  
will weigh much less.  
a. True  
*b. False  
16.4 Digester safety  
251  
46. Gas production in an anaerobic digester results from the destruction of fixed solids in the raw  
sludge.  
a. True  
*b. False  
47. "High rate" and Low rate" when used in referring to an anaerobic digester may refer to the rate of  
volatile solids loading.  
*a. True  
b. False  
48. In computing anaerobic digester loadings, it is necessary to take into account the solids lost in the  
supernatant system.  
a. True  
*b. False  
49. Increased concentrations of volatile acids and decreased alkalinity are the first measurable changes  
that take place when the process of anaerobic digestion is becoming upset.  
*a. True  
b. False  
50. A change in the-pH-of digesting sludge in anaerobic digester is the best early warning indicator of  
potential digester upset.  
a. True  
*b. False  
51. A good maintenance program should be established for all flame arrestors to ensure they are all set  
at the recommended "pop-off’ pressures.  
*a. True  
b. False  
52. A healthy anaerobic digester should have a carbon dioxide concentration of more than 40%.  
a. True  
*b. False  
53. A high rate anaerobic digester is always heated and mixed.  
*a. True  
b. False  
54. Anaerobically digested sludge produces gas as a by-product. The gas produced is of little or no  
value.  
a. True  
*b. False  
55. Digester gas containing 60% methane by volume will likely explode when exposed to a spark or  
flame.  
a. True  
*b. False  
56. Volatile solids removal efficiency of a digester is most commonly monitored utilizing the BOD test  
a. True  
*b. False  
57. Gas production in an anaerobic digester results from the destruction of fixed solids in the raw  
sludge.  
a. True  
*b. False  
252  
Chapter 16. Solids Stabilization - Digestion  
58. "High rate" and "Low rate" when used in referring to an anaerobic digester may refer to the rate of  
volatile solids loading.  
*a. True  
b. False  
59. In computing anaerobic digester loadings, it is necessary to take into account the solids lost in the  
supernatant system.  
a. True  
*b. False  
60. Increased concentrations of volatile acids and decreased alkalinity are the first measurable changes  
that take place when the process of anaerobic digestion is becoming upset.  
*a. True  
b. False  
61. In a two-stage anaerobic digestion system, it is in the secondary tank where most of the volatile  
solids destruction occurs  
a. True  
*b. False  
62. The lower explosive limit for methane is 40%  
a. True  
*b. False  
63. Volatile solids removal efficiency of a digester is most commonly monitored utilizing the BOD test  
a. True  
*b. False  
64. Sludge digestion is a process by which principally the inorganic matter in sludge is gasified and/or  
converted to a more stable form by the biological process.  
a. True  
*b. False  
65. The determination of the pH on the anaerobic digester is one of the best early warning systems of  
digester upset.  
a. True  
*b. False  
66. The gas produced as part of the anaerobic digestion is of little or no value  
a. True  
*b. False  
67. The main function of a secondary digester in normal operation is to provide a storage space for  
seed sludge.  
a. True  
*b. False  
68. The pH of digested sludge in a healthy anaerobic digester will be near 7.0.  
*a. True  
b. False  
69. Volatile solids reduction is a measure of the effectiveness of an anaerobic digester.  
*a. True  
b. False  
70. Volatile solids removal efficiency of a digester is most commonly monitored utilizing the BOD test  
a. True  
16.4 Digester safety  
253  
*b. False  
71. Identify the incorrect statement regarding anaerobic digestion.  
a. An anaerobic digester with pH of 7.05, an alkalinity of 2,900 mg/l and a volatile acid concentra-  
tion of 250 mg/l is probably operating normally.  
*b. Sodium bicarbonate may be used in place of lime to neutralize a sour anaerobic digester.  
c. When adding lime to a sour anaerobic digester, it is important to add an excess of this chemical  
to act as a reservoir of alkalinity.  
d. Ferrous sulfate may be added to an anaerobic digester to reduce hydrogen sulfide concentration  
where air quality is of concern.  
e. Gas production from anaerobic digester may be expressed as cubic feet of gas produced per  
pound of volatile matter added per day.  
72. One action that may be taken to improve the health of an anaerobic digester that is going sour  
would be to:  
a. Add small doses of lime daily to maintain the digester pH above 7.0.  
*b. Add ferrous sulfate to reduce the concentration of hydrogen sulfide in the digester.  
c. Add seed sludge from a healthy primary digester.  
d. Pump raw sludge to this digester more frequently so that this sludge has a higher pH.  
e. Increase the temperature of the digester to favor a population increase of the methane formers.  
73. Identify the incorrect statement regarding operation of an anaerobic digester.  
*a. A healthy anaerobic digester would generally have volatile acids in the range of 50 mg/l to 300  
mg/l.  
b. The best strategy for pumping raw sludge to an anaerobic digester is to pump it once or twice a  
day so that the thickest possible raw sludge may be pumped.  
c. An anaerobic digester operating at an alkalinity of 3200 mg/l should be able to tolerate a volatile  
acid concentration of 250 mg/l.  
d. The change in pH is not a reliable indicator of the changing characteristics of the digesting  
sludge because the alkalinity in an anaerobic digester acts as a buffer.  
e. The higher the volatile solid content of the primary sludge being fed to the digester, the higher is  
the expected volatile reduction.  
74. Identify the true statement about anaerobic digesters:  
a. Carbon dioxide and methane in digester gas should be 65% and 35%, respectively.  
*b. Increasing carbon dioxide readings indicate possible organic overload.  
c. Decreasing mixing will always improve recovery of a sour digester.  
d. Increasing the frequency and decreasing the amount of sludge pumping will not improve digester  
performance.  
e. Reducing the ratio of primary to waste activated sludge will improve gas production.  
75. All of the following are normal operating guidelines for a healthy anaerobic digester except for:  
a. A mesophilic digester operating at 93°F to 98°F.  
b. Methane gas in the range of approximately 62% to 70%.  
c. Carbon dioxide gas in the range of 30% to 38%.  
*d. Organic loading to a high rate digester of 0.15 to 0.2 pounds (Lb) volatile Solids per day per ft3  
of digester capacity.  
e. Bicarbonate alkalinity in the range of 1500 to 1800.  
76. When adding anhydrous ammonia to a sour primary anaerobic digester, the  
tant consideration because if it is too high, digester "poisoning" may result:  
is an impor-  
254  
Chapter 16. Solids Stabilization - Digestion  
a. Concentration of hydrogen sulfide in digester gas.  
*b. pH in digesting sludge.  
c. Concentration of free copper ions in digesting sludge  
d. Ferrous and ferric ion concentration in digesting sludge.  
e. Dissolved sulfide concentration.  
77. Which one of the following parameter ranges is most appropriate for a healthy anaerobic digester?  
a. Volatile solid reduction in the range of 60 to 70%.  
b. Digester gas production of 7 to 10 ft3 of gas produced per day per pound of volatile solids de-  
stroyed.  
*c. Alkalinity in the range of 2500 to 3500 mg/L.  
d. Hydrogen sulfide in range of 1% to 2% by volume.  
e. Volatile acids in range of 500 to 750 mg/L.  
78. Identify the incorrect statement regarding operation of an anaerobic digester.  
a. A healthy anaerobic digester would generally have volatile acids in the range of 50 mg/l to 300  
mg/l.  
*b. The best strategy for pumping raw sludge to an anaerobic digester is to pump it once or twice a  
day so that the thickest possible raw sludge may be pumped.  
c. An anaerobic digester operating at an alkalinity of 3200 mg/l should be able to tolerate a volatile  
acid concentration of 250 mg/l.  
d. The change in pH is not a reliable indicator of the changing characteristics of the digesting  
sludge because the alkalinity in an anaerobic digester acts as a buffer.  
e. The higher the volatile solid content of the primary sludge being fed to the digester, the higher is  
the expected volatile reduction.  
79. Lab data from your 100,000 gallon primary anaerobic digester, which receives primary sludge only,  
is shown below. Using this data :  
Date  
pH  
Alkalinity(mg/L)  
3,200  
Vol.Acids(mg/L)  
CO2(%)  
35.5  
9/02  
7.10  
280  
9/09  
9/16  
9/17  
7.00  
6.90  
6.85  
3,020  
2,800  
2,720  
320  
400  
450  
36.0  
37.7  
38.2  
Date  
RawSludge(TS%)  
5.4  
Raw Sludge(VS%)  
65.5  
Digested Sludge(VS%)  
56.0  
9/02  
9/09  
9/16  
5.0  
4.9  
66.7  
65.9  
53.8  
54.2  
(a) Calculate the average volatile solids reduction. Compare your calculated value to generally  
accepted ranges for a healthy anaerobic digester. Comment.  
(b) Compare the other data to expected ranges.  
(c) Is this digester experiencing an operational problem ? If so, what is the problem. Name three  
16.4 Digester safety  
steps that may be taken to mitigate the problem.  
255  
(d) Should slake lime be added ? Why or why not ?  
80. As part of an ongoing training program at your wastewater plant you are assigned the task of  
preparing a lecture for OITs on the subject of "buffer" in an anaerobic digester. Answer the fol-  
lowing questions in preparation for your presentation.  
a. What is "buffer" and why is it important in an anaerobic digester?  
b. What lab test is used to measure "buffer" in an anaerobic digester? Explain how this test is  
run  
c. Most of the buffer in an anaerobic digester is due to the presence of  
17. Solids Dewatering  
Dewatering, like thickening does not treat the sludge but it allows for a reduction in sludge volume  
by removing water.  
Thickening achieves about 10% or less solids content while dewatering is typically for increasing  
the solids content to between 15 to 30 percent.  
Sludge is dewatered to make it easier to handle and to reduce costs associated with elements re-  
lated to accomplishing the end objectives with the sludge – land application, composting, drying,  
incineration or landfill.  
Dewatering involves conditioning the sludge with a polymer and subjecting it to a physical process  
such as belt filter press or centrifuge to remove the water.  
• Reasons for sludge dewatering are:  
Critical for sludge treatment options such as sludge drying and incineration.  
Reduction in the weight of solids to be hauled – reducing hauling cost.  
Reduction in the volume of sludge that needs to be handled  
• More common sludge dewatering methods include:  
1. Belt Filter Press  
2. Centrifuge  
• These are both mechanical methods and involve physically removing free water  
Both these methods involve conditioning of sludge with a cationic polymer which flocculate the  
solids - separating it from the water  
17.1 Belt Filter Press  
The belt filter press dewaters sludge by squeezing polymer conditioned sludge between a pair of belt  
filter fabric as these belts pass through a system of rollers. Belt filter press produces a dewatered product  
typically between 12% to 35% solids content.  
 
 
258  
Chapter 17. Solids Dewatering  
Belt Filter Press  
17.1.1 Principles of belt filter press operations  
Belt filter press utilizes two endless, porous belts (generally 0.7 to 3 meter wide) made from a  
synthetic material  
Polymer flocculated sludge is first introduced on the horizontal gravity zone of the press where free  
water is removed from the sludge by chicane assisted gravity filtration through the porous belt press  
fabric.  
 
17.1 Belt Filter Press  
259  
Gravity Zone - Chicanes  
From the gravity zone, the sludge enters the low pressure zone (wedge zone) in which the sludge is  
prepared for the upcoming high-pressure zone by evenly distributing the solids across the belt and  
gradually applying pressure  
In the high pressure zone the sludge is sandwiched between two belt press fabrics. As the belts  
with the sandwiched sludge passes over and under 6 to 12 tensioning rollers which progressively  
decrease in diameter. The applied pressure squeezes the water out from the sludge and is removed  
through the porous fabric.  
• The filtrate is returned back to the front of the plant.  
The belts are washed continuously as they pass over wash boxes. The wash boxes are equipped with  
rotating brushes and water sprays and are primarily for keeping the belt press fabric pores open so  
water could easily filter through.  
• Improper polymer dosage and excessive hydraulic loading can lead to a wetter sludge cake.  
The belts are operated at speeds that commensurate with the rate of solids loading and the type  
of sludge being dewatered. Lower belt speed would allow for better water removal. However, if  
if the belts speed is below the threshold for that particular sludge, belt press washout, which is  
characterized by sludge flowing over the sides of the belt can occur . Thus, the best belt speed is the  
slowest speed the belt can be operated without causing washout.  
The belts can be blinded - loose its porosity causing the water to stagnate/cause washout and not  
filter through, if excessive polymer is used or due ineffective belt washing.  
Solids capture rate and Percent cake solids produced are the most important belt press efficiency  
measurement parameters  
17.1.2 Elements of the belt filter press  
• belts  
• wash boxes  
• guiding and tensioning rollers  
• chicanes  
• doctor blades  
• drive motor  
• hydraulic unit  
• polymer injection and mixing system  
 
260  
Chapter 17. Solids Dewatering  
17.1.3 Belt press operational parameters  
• belt filter width  
• belt filter speed  
• hydraulic loading  
• belt tension  
• washout  
• filtrate quality  
• solids capture rate  
• polymer dosing rate (lbs polymer/dry ton solids)  
• solids and hydraulic loading rates, and  
• quantity and characteristics of the polymer used  
17.2 Centrifuge  
The centrifuge dewaters the sludge by subjecting a polymer conditioned sludge to strong centrifugal  
forces by rotating it in a bowl at high speeds. Centrifuges are used for dewatering solids in many differ-  
ent applications and wastewater solids dewatering being one them. Centrifuges also come in different  
configurations. The scroll conveyor (decanter) centrifuge is the more commonly used centrifuge design  
for wastewater sludge dewatering. It produces a dewatered product typically between 20% to 30% solids  
content. It should be noted that the centrifuge can also be utilized for sludge thickening.  
Centrifuge  
17.2.1 Principles of centrifuge operations  
The bowl of the centrifuge is cylindrical shaped with a tapered cone at one end. The bowl is de-  
signed to rotate at speeds of 1200 to 2400 RPM.  
 
 
 
17.2 Centrifuge  
261  
Centrifuge Bowl  
A screw conveyor - scroll is shaped to the bowl contour and carried on a central shaft or drum and  
it rotates independently from the bowl.  
Centrifuge Scroll  
The polymer conditioned sludge is fed into the cylindrical portion of the bowl through a central  
feed pipe. Inside the bowl, the sludge is subjected to intense G forces around 3000 Gs by virtue of  
the bowl rotating at a high speed.  
• As the bowl rotates, the water in the flocculated sludge is separated from the solids .  
The scroll rotates at a 5 to 15 rpm differential speed from the bowl. This differential speed causes  
the thickened sludge to be propelled along the bowl for towards its conical end. As the sludge is  
screwed up the cone it emerges from the liquid at the beach, is drained and pushed up to discharge  
ports.  
The separated water is discharged from the cylindrical (non-conical) end of the bowl. The depth of  
the water - pond depth, is controlled by adjustable weirs at the discharge end.  
The greater the pond depth improves the centrate quality and solids recovery but it reduces the  
drainage zone at the beach end resulting in wetter solids. The pond depth is altered using adjustable  
weirs.  
• The centrate is returned back to the front of the plant  
17.2.2 Elements of the Centrifuge  
• bowl  
• scroll  
• beach  
 
262  
Chapter 17. Solids Dewatering  
• adjustable weir  
• differential gear box  
• drive motor  
17.2.3 Centrifuge operational parameters  
• bowl speed  
• differential speed  
• pond height  
• centrate quality  
• solids capture rate  
• polymer dosing rate (lbs polymer/dry ton solids)  
• solids and hydraulic loading rates  
 
Chapter Assessment  
1. What can be a problem with a belt filter press?  
a. Washing out  
b. Polymer overdosing  
c. Blinding  
*d. All of the above  
2. Which one of the following statements is TRUE in regard to the operation of a belt filter press:  
a. Unlike a gravity belt, cationic polymers need not be used to condition the feed sludge to this  
unit.  
b. When dewatering a mixture of WAS and primary sludge, which has been anaerobically digested,  
the operator can expect to produce a sludge cake with 22% TS to 25%.  
*c. The best belt speed for this unit is the slowest speed the belt can be operated at, without causing  
“washout.”  
d. Colloidal solids are likely to cause plugging of belt pores and thus cause “belt binding.”  
3. What can be used to evaluate the efficiency of a belt filter press?  
a. Vacuum required in inches of mercury  
b. % volatile solids in cake  
c. Sludge feed rate in gpd  
*d. Filter yield in lbs/hr/sq ft  
e. Gph of filtrate removal.  
4. An increase in the pool depth of a scroll-type centrifuge:  
a. would not affect the moisture content of the cake.  
b. would produce a drier cake  
*c. would produce a wetter cake, but produce a greater solids recovery.  
d. would not affect either solids recovery, nor cake moisture content.  
e. would require an increase in the cationic polymer dosage.  
5. What can be used to evaluate the efficiency of a belt filter press?  
264  
Chapter 17. Solids Dewatering  
a. Vacuum required in inches of mercury  
b. % volatile solids in cake  
c. Sludge feed rate in gpd  
*d. Filter yield in lbs/hr/sq ft  
e. gallons-per-hour of filtrate removal.  
6. You are assigned to supervise the operation ofa belt filter press. Anerobically digested sludge - a  
mixture of primary and secondary sludge is being dewatered. Lab records show that this belt press  
routinely produces a sludge cake having a total solids concentration of 16.5%. No other lab tests  
are run on the belt press. Answer the following questions:  
a. Are you satisfied with the dryness of the cake? Why or why not? Explain your answer.  
b. List three (3) variables you could adjust to optimize the BFP. Explain how these adjustments  
affect the optimization of this unit.  
c. List two (2) other lab tests that must be performed in order to better monitor the operation,of  
the BFP  
18. Disinfection  
18.1 Background  
The primary goal of water treatment is to ensure that the water is safe to drink and does not contain  
any disease-causing microorganisms.  
Disinfection refers to an operation to inactivate the microorganisms in water that can cause an  
infection or disease. These organisms are collectively referred to as pathogens and include many  
species of bacteria, fungus, protozoa, worms, viruses, etc.  
The processes prior to disinfection - sedimentation and filtration, remove a large percentage of  
bacteria and other microorganisms from the water by physical means.  
Disinfection is different from sterilization, which is the complete destruction of all organisms  
which is expensive and unnecessary.  
• Water disinfection can be sub-divided as:  
1. Primary disinfection:  
Kills or inactivates bacteria, viruses, and other potentially harmful organisms in drinking  
water.  
Disinfection prevents infectious diseases such as typhoid fever, hepatitis, and cholera  
Some disinfectants are more effective than others at inactivating certain potentially  
harmful organisms.  
Disinfection processes vary from water utility to water utility based on their needs and  
to meet EPA treatment requirements.  
2. Secondary disinfection:  
Maintenance of a disinfectant residual that prevents regrowth of microorganisms in the  
water distribution system between treatment and consumer.  
Secondary disinfection maintains water quality by killing potentially harmful organisms  
such as those that cause Legionnaire’s disease that may get in water as it moves through  
pipes.  
 
 
266  
Chapter 18. Disinfection  
Monochloramine is commonly used as a secondary disinfectant.  
• Elements of an "ideal" disinfectant  
It must act in a reasonable time.  
It must act as temperature or pH changes.  
It must be nontoxic.  
No harmful byproducts.  
It must not add unpleasant taste or odor.  
It must be readily available.  
It must be safe and easy to handle and apply.  
It must be easy to determine the concentration of.  
It must be able to provide residual protection.  
Pathogenic organisms must be more sensitive to the disinfectant than are nonpathogens.  
It must be capable of being applied continually.  
Versatile: effective against all types of pathogens.  
Fast-acting: effective within short contact times  
Robust: effective in the presence of interfering materials including particulates, suspended  
solids and other organic and inorganic constituents  
Handy: easy to handle, generate, and apply (nontoxic, soluble, non-flammable, non-explosive)  
Compatible with various materials/surfaces in WTPs (pipes, equipments)  
Economical  
In addition to the desirable characteristics of a disinfectant listed above, the disinfectant chosen  
must be able to kill off or deactivate pathogenic microorganisms by one of several possible meth-  
ods, including:  
1. Damaging the cell wall  
2. Altering the ability to pass food and waste through the cell membrane  
3. Altering the cell protoplasm  
4. Inhibiting the cells’ conversion of food to energy  
5. Inhibiting reproduction  
18.2 Chlorination  
• Despite potential drawbacks, chlorine is the disinfectant of choice.  
In general, chlorination is effective, relatively inexpensive, and provides effective levels of disinfec-  
tant residual for safe distribution.  
• Chlorine can be applied as:  
As a gas - elemental chlorine, Cl2:  
Liquid (sodium hypochlorite)  
Solid (calcium hypochlorite)  
each of these forms has advantages and disadvantages.  
18.2.1 Chlorine properties  
• Chlorine is a yellowish-green gas at room temperature and atmosphric pressure  
• Chlorine gas can be pressurized and cooled to its liquid form for making it easy to ship and store.  
When liquid chlorine is released, it quickly turns into a gas that stays close to the ground (being  
heavier than air) and spreads rapidly.  
 
 
18.2 Chlorination  
267  
While it is not explosive or flammable, as a liquid or gas it can react violently with many sub-  
stances  
• Chlorine is only slightly soluble in water (0.3 to 0.7% by weight.)  
• Chlorine gas has a greenish-yellow color  
It has a characteristic disagreeable and pungent odor, similar to chlorine-based laundry bleaches,  
and is detectable by smell at concentrations as low as 0.2 to 0.4 ppm  
• It is about two and a half times as heavy as air  
• One volume of liquid chlorine yields about 460 volumes of chlorine gas.  
• Liquid chlorine is amber in color and is about one and a half times as heavy as water  
• Chlorine is an irritant to the eyes, skin, mucous membranes, and the respiratory system  
18.2.2 Chlorine storage and safety  
Chlorine gas is lethal at concentrations as low as 0.1% air by volume. In nonlethal concentrations,  
it irritates the eyes, nasal membranes, and respiratory tract.  
Typically for smaller plants chlorine gas is shipped in pressurized steel cylinders - 150 lb or 2000 lb  
(ton cylinder) size. Larger plants may get their chlorine supply in rail tank cars.  
The daily chlorine usage is typically established based upon the weighing of the chlorine contain-  
ers.  
The withdrawal rates from a chlorine cylinder is based on the temperature of the liquid in the  
cylinder, and thus the pressure of the gas.  
• As chlorine gas is withdrawn from the cylinder, it absorbs the heat from the surroundings.  
For low withdrawal rates, heat will be able to be transferred from the surrounding air to the con-  
tainer in time so that there is no drop in temperature or pressure,  
If the chlorine withdrawal is larger, the air will not be able to transfer the heat quickly enough and  
the temperature (and pressure) of the chlorine will drop, thus resulting in a lower feed rate.  
If high enough and prolonged enough, this can even result in ice formation around the outside of  
the container, further decreasing the withdrawal rate.  
The most effective way to increase withdrawal rate from a single container is to circulate the sur-  
rounding air with a fan. Again, never apply heat to the containers.  
If chlorine gas escapes from a container or system, being heavier than air, it will seek the lowest  
level in the building or area  
Only trained staff with access to proper personal protection equipment (PPE) including self-  
contained breathing apparatus, should handle the chlorine cylinders and address chlorine leak  
issues  
When a leak is suspected, it is recommended that ammonia vapors be used to find the source. When  
ammonia vapor using a rag or brush, is directed at a leak, a white cloud will form. To produce am-  
monia vapor, a plastic squeeze bottle containing about 5 % ammonia, aqua ammonia (ammonium  
hydroxide solution) should be used. A weaker solution such as household ammonia may not be  
concentrated enough to detect minor leaks  
All safety equipment should be located outside of the chlorine room and be easily accessed by all  
personnel  
Small leaks around valve stems can usually be corrected by tightening the packing nut or closing  
the valve. A leak can also be reduced by removing the chlorine as rapidly as possible  
If it cannot be added to the process there are several chemicals which can be used to absorb the  
chlorine gas. For example, chlorine can be absorbed by using 1frac14 pounds of caustic soda or  
 
268  
Chapter 18. Disinfection  
hydrated line, or 3 pounds of soda ash per pound of chlorine.  
If the leaking container can be moved, it should be transported to an outdoors area where minimal  
harm will occur. Keep the leaking part the most elevated so that gaseous chlorine will leak rather  
than liquid chlorine.  
If the leak is large, all persons in the adjacent area must be warned and evacuated. Only authorized  
persons equipped with the proper breathing apparatus, and protective measures to the eyes and  
body should investigate.  
As water is not an efficient absorbent for chlorine and the fact that chlorine reacts with water to  
form very corrosive hydrochloric acid, never apply water to a leak or consider submerging a chlo-  
rine cylinder (for example, in a pond or tank), since it will probably float.  
• Remember to keep windward of the leak.  
As chlorine cylinders pressure increases with temperature, as a safety measure the chlorine cylin-  
ders are fitted with fusible plug which melts between 158o and 165o F.  
Keep chlorine cylinder or container emergency repair kits available. Be familiar with their use and  
location.  
Leaks at fusible plugs and cylinder valves requires special handling and emergency equipment. The  
chlorine supplier must be notified immediately  
Pin hole leaks in cylinder walls or ton tanks can usually be stopped by mechanical pressure applica-  
tions (clamps, turnbuckles, etc.). This only temporary and may require your ingenuity.  
• Leaking containers cannot be shipped.  
• In general, daily inspection of all chlorine cylinders will avoid major problems  
18.2.3 Forms of chlorine  
Due to safety issues related to the use of chlorine gas, hypochlorites are often used in lieu of  
chlorine  
• Types of hypochlorites  
Sodium hypochlorite (NaOCl) comes in a liquid form which contains up to 12.5% chlorine  
Calcium hypochlorite (Ca(OCl)2), also known as High-test Hypochlorite (HTH), is a solid  
which is mixed with water to form a hypochlorite solution. Calcium hypochlorite is 65-70%  
concentrated.  
Hypochlorites decompose in strength over time while in storage. Temperature, light, and physical  
energy can all break down hypochlorites before they are able to react with pathogens in water.  
18.2.4 Chlorine reactions related to disinfection  
Chlorine reacts with water to form hypochlorous and hydrochloric acids  
Cl2  
+
H2O  
⇐⇒  
HOCl  
+
HCl  
chlorine  
water  
hypochlorous acid hydrochloric acid  
• Hypochlorous acid dissociates in water to form the hydrogen and hypochlorite ions  
HOCl  
⇐⇒  
H+  
+
OCl−  
hypochlorous acid  
hydrogen ion  
hypochlorite ion  
Hypochlorous acid is the most effective form of chlorine available to kill microorganisms  
Hypochlorite ions is much less efficient disinfectant  
The concentration of hypochlorous acid and hypochlorite ions in chlorinated water will depend on  
the water’s pH  
 
 
18.2 Chlorination  
269  
A higher pH facilitates the formation of more hypochlorite ions and results in less hypochlor-  
ous acid in the water  
A significant percentage of the chlorine is still in the form of hypochlorous acid even between pH 8  
and pH 9  
18.2.5 Chlorine disinfection process  
When chlorine is added to a wastewater flow, it will first react or combine with certain organic and  
inorganic substances present, prior to acting on pathogens. The amount of chlorine used up as part  
of these reactions is referred to as the chlorine demand  
The free chlorine remaining after the chlorine demand is satisfied, is the strongest form of chlorine  
available for disinfection.  
Chlorine combined with ammonia (as chloramines) and organic compounds (as chloroorganic  
compounds), known as combined chlorine also exhibit disinfecting properties - albeit weaker than  
the free chlorine.  
• Total residual chlorine is the sum of free chlorine and combined chlorine and it is the residual  
chlorine concentration which represents the amount of chlorine available for disinfection  
Chlorine Demand = Applied Chlorine Dose - Chlorine Residual  
18.2.6 Factors affecting chlorine disinfection  
The disinfection efficiency of chlorine depends on the following factors:  
pH: Disinfection is more efficient at a low pH when large quantities of hypochlorous acid are  
present than at a high pH when hypochlorite ions is the dominant species in the water  
Concentration: Contact Time Ratio (CT): For effective chlorine disinfection both sufficient chlorine  
dosages – concentration (C) as well as contact time (T) are necessary. Generally both of these  
factors must be worked out experimentally for a given system  
Disinfection activity can be expressed as the product of disinfection concentration (C) and contact  
time (T) - CT  
• The same CT values will achieve the same amount of inactivation  
Temperature: Colder temperatures are less favorable for disinfection. Proper contacting or mix-  
ing or agitation: This is necessary to make sure that the chlorine applied contacts or reaches the  
microbial cells  
Organic and inorganic material present: The chlorine used by these organic and inorganic reducing  
substances including metal ions, organic matter and ammonia, is defined as the chlorine demand.  
So that the amount of chlorine that has to be added to wastewater for different purposes will also  
vary.  
18.2.7 Chlorine Application  
Continuous chlorination of systems less than 75 gpm is by the use of a hypochlorinator where a  
motor driven pump pulls the hypochlorite solution out of a holding chamber and pumps it into  
the water to be treated. Where the pipe from the pump joins the pipe carrying the raw water, the  
Venturi effect creates a small vacuum and pulls the chlorine solution into the water.  
For larger system, chlorinators - devices which introduce chlorine gas to water using liquid chlorine  
supplied in steel cylinders are used.  
• In the commonly used Vacuum Chlorinator:  
 
 
 
270  
Chapter 18. Disinfection  
Chlorine gas is pulled from the cylinder into the source water by a vacuum created by water  
flowing through the injector and creating a negative head.  
This negative head forces open the pressure regulating valve on the cylinder and allows  
chlorine gas to flow out of the cylinder and into the chlorinator.  
Once the gas has entered the chlorinator, the chlorine feed rate is measured using an indicator  
known as a rotameter  
Just beyond the rotameter, the chlorine gas flows past a regulating device (a V-notch plug or a  
valve) which is used to adjust the chlorine feed rate.  
Figure 18.1: Vacuum Chlorinator  
18.2.8 Chlorination Byproducts  
Main drawback of chlorine disinfection is the adverse health effects of the byproducts - Disinfec-  
tion by-products ( DBPs ) formed from its reaction with certain organic compounds present in the  
water.  
Adverse health effects on humans exposed to DBPs through drinking-water and oral, dermal, and  
inhalational contact with chlorinated water, include cancers of vital organs.  
Halogenated trihalomethanes ( THMs ) and haloacetic acids ( HAAs ) are two major classes of  
disinfection byproducts (DBPs) commonly found in waters disinfected with chlorine.  
At the present time, about 90% of U.S. water utilities use chlorine to disinfect water. Although  
chlorine has virtually eliminated the risks of waterborne disease such as typhoid fever, cholera, and  
dysentery, recent studies have shown risks associated with byproducts of chlorine—a reason why  
water utilities already have been looking at alternative methods for disinfecting water.  
• Approaches for reducing DBPs includes:  
Avoiding pre-chlorination where chlorine is added to the raw water before coagulation and  
filtration.  
Removal of organics using Aeration or adsorption on activated carbon  
reevaluating the chlorine dosing to a level which will accomplish the same degree of disinfec-  
tion with a lower chlorine dosage.  
Another current approach is using alternative disinfection methods.  
 
18.2 Chlorination  
271  
18.2.9 Chloroamination  
When chlorine is added to water containing ammonia, chlorine reacts with ammonia to form chlo-  
ramines.  
Chloramines has disinfection properties and although it is a weaker disinfectant than chlorine,  
it is more stable enabling to extend its disinfectant benefits through a wider range of the water  
distribution system.  
Additionally, water disinfection using chloramines provides benefit related to fewer taste and odor  
issues when compared to water disinfected with chlorine.  
Water utilities practice chloroamination - practice of utilizing the disinfectant properties of chlo-  
ramines, by feeding chlorine to the water containing ammonia.  
• When chlorine is added to water containing ammonia, the :  
an initial absence of any increase in combined chlorine residual  
followed by a decrease in the combined chlorine residual along with ammonia concentrations  
followed by an increase in free chlorine residual and near complete removal of ammonia as  
nitrogen gas.  
Chlorine reacts with ammonia to form chloramines  
1. First the free chlorine in contact with ammonia forms monochloramine and water  
Monochloramine has disinfection properties  
*
*
*
Dominates when Cl:N mass ratio is 0 to 5:1  
The breakpoint curve rises at about 1:1 during monochloramine formation  
NH3 +HOCl NH2Cl(monochloramine)+H2O  
2. Monochloramine reacts further with chlorine to give dichloramine and water  
HOCl +NH2Cl NHCl2 +H2O  
Also, monochloramine auto decomposes into dichloramine  
2NH2Cl NHCl2 +NH3  
Between dichloramine is formed between 5:1 and 7:1 Cl:N mass ratio  
When you are getting significant dichloramine, the breakpoint curve will start  
dropping  
*
*
NH2Cl +HOCl NHCl2(dichloramine)  
at pH 7.5, monochloramine is the dominant chloramine species as pH decreases from  
>
7.5, dichloramine becomes the dominant chloramine species increases in the chlorine to  
nitrogen dose ratio results in corresponding increases of nitrogen trichloride, but only  
when the pH is < 7.4  
3. Formation of nitrogen trichloride from the reaction of chlorine and dichloramine does  
not typically occur as it is the favored product at low pH - <4  
NHCl2 +HOCl NCl3(nitrogentrichloride)  
4. Additional free chlorine +chloramines H+ +H2O+N2  
Chloramine levels up to 4 milligrams per liter (mg/L) or 4 parts per million (ppm) are consid-  
ered safe in drinking water. At these levels, harmful health effects are unlikely to occur.  
Chloramines have disinfection properties albeit much lower than free chlorine ( 5% of free  
available chlorine) but last much longer in the system than free chlorine.  
Monochloramine is about 2,000 and 100,000 times less effective than free chlorine for the  
inactivation of E. Coli and rotaviruses, respectively.  
After the breakpoint, free chlorine residuals develop. Free chlorine residuals usually destroy  
odors, kill microorganisms and oxidize organic matter.  
 
272  
Chapter 18. Disinfection  
Breakpoint chlorination is the application of sufficient chlorine beyond the chlorine demand  
to maintain a free available chlorine residual.  
Theoretically chlorine requirement = Wt. NH3-N x 7.6  
in practice (Margin of safety) = Wt. NH3-N x 10  
Thus, breakpoint chlorination is possible if ratio of Cl2 to ammonia exceeds 10:1 then free  
Cl2 may exist. Cl2 demand will be high because the reaction of free Cl2 with nitrite and other  
organic compounds.  
Theoretically, while microorganisms are killed as the chlorine demand is being satisfied,  
disinfection is generally the result of chlorine residual or the amount of chlorine remaining  
after the chlorine demand has been satisfied.  
The following chart is a graphical explaination of the concept of Breakpoint Chlorination  
Point A is at the beginning of chlorine application  
Between Points A and B, the chlorine dosage produces no residual because of an immediate  
chlorine demand caused by fast-reacting ions from metal salts and H2S.  
Point B is the beginning of the reaction between chlorine and ammonia present  
Mono and dichloramines are formed between points B and C  
Zone 1 - between points A and C, is the combined zone and has mono and di chloramines  
and ammonia. Mono chloramines is a stable disinfectant while dichloramines is a strong  
disinfectant but unstable.  
After the maximum combined residual is reached (point C), further chlorine doses decrease  
the residual due to chloramine oxidation to dichloramine, occurring between points C and D.  
18.2 Chlorination  
This is Zone 2 - Breakpoint Zone  
273  
Point D represents the breakpoint - the point at which chlorine demand has been satisfied and  
additional chlorine appears as free residuals  
Between points D and E, free available residual chlorine increases in direct proportion to the  
amount of chlorine applied. This is Zone 3 which is the free chlorine zone and has hypochlor-  
ous acid but no ammonia.  
Factors that affect breakpoint chlorination are initial ammonia nitrogen concentration, pH,  
temperature, and demand exerted by other inorganic and organic species  
Weight ratio of chlorine applied to initial ammonia nitrogen must be 8:1 or greater for the  
breakpoint to be reached. If the weight ratio is less than 8:1, there is insufficient chlorine  
present to oxidize the chlorinated nitrogen compounds initially formed  
When instantaneous chlorine residuals are required, the chlorine needed to provide free avail-  
able chlorine residuals may be 20 or more times the quantity of ammonia present. Reaction  
rates are fastest at pH 7-8 and high temperatures  
18.2.10 Chlorine dosing terms  
Chlorine dose - the amount of chlorine added to the system. It can be determined by adding the  
desired residual for the finished water to the chlorine demand of the untreated water. Dosage can be  
either milligrams per liter (mg/L) or pounds per day (lb/day).  
Chlorine Demand - the amount of chlorine consumed by iron, manganese, turbidity, algae, and  
microorganisms in the water. Because the reaction between chlorine and microorganisms is not in-  
stantaneous, demand is relative to time. For instance, the demand 5 minutes after applying chlorine  
will be less than the demand after 20 minutes.  
Free chlorine - free chlorine refers to all chlorine present in the water as Cl2(g), HOCl(aq) and  
OCl(aq).  
Combined residual - is the result of combining free chlorine with nitrogen compounds. Combined  
residuals are also referred to as chloramines.  
Total chlorine residual - is the mathematical combination of free chlorine and combined residuals.  
Total residual can be determined directly with standard chlorine residual test kits. Residual, like  
demand, is based on time. The longer the time after dosage, the lower the residual will be, until  
all of the demand has been satisfied. Residual, like demand, is expressed in mg/L. The presence  
of a free residual usually provides a high degree of assurance that the disinfection of the water is  
complete.  
ChlorineDose(mg/L) = ChlorineDemand+ChlorineResidual  
• Demand, like dosage, is expressed in mg/L. The chlorine demand is as follows:  
18.2.11 Chlorine dosing problems  
Example 1:  
Determine the chlorinator setting (lb/day) required to treat a flow of 4MGD with a chlorine dose of  
5mg/L.  
Chlorine feed rate (lb/ day ) = Chlorine (mg/L)× Flow (MGD)×8.34lb/gal  
Chlorine feed rate (lb/ day ) = 5mg/L×4MGD×8.34lb/gal  
Chlorine feed rate (lb/ day ) = 167lb/ day  
 
 
274  
Chapter 18. Disinfection  
Example 2 :  
A pipeline that is 12 inches in diameter and 1400ft long is to be treated with a chlorine dose of 48mg/L  
.
How many lb of chlorine will this require?  
First determine the gallon volume of the pipeline:  
Volume (gal) = 0.785×D2× length (ft)×7.48gal/cuft  
Volume (gal) = 0.785×(1ft)2 ×1400ft×7.48gal/cuft Volume (gal) = 8221gal  
Next calculate the amount of chlorine required:  
Chlorine feed rate (lb/ day ) = Chlorine (mg/L) x Flow ( MGD) ×8.34lb/gal  
Chlorine feed rate (lb/ day ) = 48mg/L×0.008221MGD×8.34lb/gal  
Chlorine feed rate (lb/ day ) = 3.3lb  
Example 3:  
A water sample is tested and found to have a chlorine demand of 1.7mg/L. If the desired chlorine residual  
is 0.9mg/L, what is the desired chlorine dose (in mg/L )?  
Chlorine Dose (mg/L) = Chlorine Demand + Chlorine Residual  
Chlorine Dose (mg/L) = 1.7mg/L+0.9mg/L  
Chlorine Dose(mg/L) = 2.6mg/L  
Example 4:  
The chlorine dosage for water is 2.7mg/L. If the chlorine residual after a 30-minute contact time is found  
to be 0.7mg/L, what is the chlorine demand (in mg/L )?  
Chlorine Demand = Chlorine Dose Chlorine Residual  
Chlorine Demand = 2.7mg/L0.7mg/L  
Chlorine Demand = 2.0mg/L  
Example 5:  
What should the chlorinator seting be (lb/day) to treat a flow of 2.35MGD if the chlorine demand is  
3.2mg/L and a chlorine residual of 0.9mg/L is desired?  
First, determine the chlorine dosage (in mg/L ):  
Chlorine Dose (mg/L) = Chlorine Demand + Chlorine Residual  
Chlorine Dose (mg/L) = 3.2mg/L+0.9mg/L  
Chlorine Dose (mg/L) = 4.1mg/L  
Next calculate the chlorine dosage (feed rate) in lb/ day:  
Chlorine feed rate (lb/ day ) = Chlorine (mg/L)× Flow (MGD)×8.34lb/gal  
Chlorine feed rate (lb/ day ) = 4.1mg/L×2.35MGD×8.34lb/gal  
Chlorine feed rate (lb/ day ) = 80.4lb/ day  
Example 6:  
A chlorinator setting is increased by 2lb/ day. The chlorine residual before the increased dosage was  
0.2mg/L. After the increased chlorine dose, the chlorine residual was 0.5mg/L. The average flow rate  
being chlorinated is 1.25MGD. Is the water being chlorinated beyond the breakpoint?  
First calculate the expected increase in chlorine residual:  
Chlorine feed rate (lb/ day ) = Chlorine (mg/L) x Flow (MGD)×8.34lb/gal  
2lb/ day = ×mg/L×1.25MGD×8.34lb/gal  
x = 2/(1.25×8.34)  
x = 0.19mg/L  
Actual increase in residual is:  
0.5mg/L0.19mg/L = 0.31mg/L  
Example 7:  
18.2 Chlorination  
275  
A chlorinator setting of 18lb chlorine per 24 hours result in a chlorine residual of 0.3mg/L. The chlori-  
nator setting is increased to 22lb per 24 hours. The chlorine residual increased to 0.4mg/L at this new  
dosage rate. The average flow being treated is 1.4MGD. On the basis of these data, is the water being  
chlorinated past the breakpoint?  
First calculate the expected increase in chlorine residual:  
Chlorine feed rate (lb/ day ) = Chlorine (mg/L)× Flow (MGD)×8.34lb/gal  
4lb/ day = xmg/L×1.4MGD×8.34lb/gal  
x = 4/(1.4×8.34)  
x = 0.34mg/L  
Next calculate the actual increase in residual:  
0.4mg/L0.3mg/L = 0.1mg/L  
Chapter Assessment  
1. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports  
even when empty. Except when actually being moved to or from storage.  
*a. True  
b. False  
2.  
3. Chlorine demand is a good indicator of effective disinfection  
a. True  
*b. False  
4. Chlorine demand is a good indicator of effective disinfection  
a. True  
*b. False  
5. Chlorine demand is defined as the amount of chlorine remaining in the waste water at the end of a  
specific contact period.  
a. True  
*b. False  
6. Chlorine disinfection is more effective at higher pH  
a. True  
*b. False  
7. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the  
amount of residual chlorine remaining after a given contact time.  
a. True  
*b. False  
8. Chlorine feed rate and chlorine residual are both usually expressed in units of ppm.  
*a. True  
b. False  
9. Chlorine is used to sterilize wastewater in order to insure protection of public health.  
278  
Chapter 18. Disinfection  
a. True  
*b. False  
10. Chlorine demand is a good indicator of effective disinfection  
a. True  
*b. False  
11. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air  
a. True  
*b. False  
12. Chlorine demand is defined as the amount of chlorine remaining in the waste water at the end of a  
specific contact period.  
a. True  
*b. False  
13. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the  
amount of residual chlorine remaining after a given contact time.  
a. True  
*b. False  
14. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.  
a. True  
*b. False  
15. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports  
even when empty. Except when actually being moved to or from storage.  
*a. True  
b. False  
16. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air  
a. True  
*b. False  
17. Chlorine demand is defined as the amount of chlorine remaining in the waste water at the end of a  
specific contact period.  
a. True  
*b. False  
18. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the  
amount of residual chlorine remaining after a given contact time.  
a. True  
*b. False  
19. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.  
a. True  
*b. False  
20. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports  
even when empty. Except when actually being moved to or from storage.  
*a. True  
b. False  
21. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the  
amount of residual chlorine remaining after a given contact time.  
a. True  
*b. False  
18.2 Chlorination  
279  
22. Before any new or repaired chlorine piping system is placed into operation, it should be pressure  
tested with water to detect any possible leaks before chlorine gas or liquid is used.  
a. True  
*b. False  
23. To produce a savings in chlorination cost and power consumption, chlorine feed rates for odor  
control should be determined at low flow conditions.  
*a. True  
b. False  
24. Chlorine is used to sterilize wastewater in order to insure protection of public health.  
a. True  
*b. False  
25. The purpose of disinfection with chlorine is to destroy pathogenic organisms.  
*a. True  
b. False  
26. Oxidation pond effluents are easily disinfected with chlorine because of the large amount of algae  
present  
a. True  
*b. False  
27. The chlorine demand of wastewater equals the chlorine dosage plus the residual.  
a. True  
*b. False  
28. Post chlorination is employed primarily for odor control and BOD reduction.  
a. True  
*b. False  
29. The safety plugs on a one-ton chlorine tank are designed to soften or melt at temperatures in excess  
of 158°F.  
*a. True  
b. False  
30. When moving a chlorine cylinder a short distance, for example 10 feet, it is not necessary to replace  
the protective cap.  
a. True  
*b. False  
31. A leaking chlorine cylinder cannot be transported.  
*a. True  
b. False  
32. Canister-type gas masks are not recommended for use when working ·on chlorine leaks, because  
they do not supply oxygen to the wearer.  
*a. True  
b. False  
33. Canister-type gas masks may be used to attend chlorine gas leaks at concentrations in the range of  
two to five percent.  
a. True  
*b. False  
34. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.  
a. True  
280  
Chapter 18. Disinfection  
*b. False  
35. Objective of the disinfection process is to sterilize the wastewater  
a. True  
*b. False  
36. Caution needs to be exercised when handling chlorine due to its explosive properties  
a. True  
*b. False  
37. It is important that the (PPE) including self-contained breathing apparatus, for handling leaks from  
chlorine cylinders is located inside the chlorine storage building to ensure easy access  
a. True  
*b. False  
38. Ammonia solution is used for finding leaks in chlorine cylinders  
*a. True  
b. False  
39. Sodium hypochlorite is more effective and less expensive than chlorine gas  
a. True  
*b. False  
40. Sodium hypochlorite is more effective and less expensive than chlorine gas  
a. True  
*b. False  
41. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air  
a. True  
*b. False  
42. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports  
even when empty. Except when actually being moved to or from storage.  
*a. True  
b. False  
43. Chlorine demand is a good indicator of effective disinfection  
a. True  
*b. False  
44. Chlorine demand is a good indicator of effective disinfection  
a. True  
*b. False  
45. Chlorine demand is defined as the amount of chlorine remaining in the waste water at the end of a  
specific contact period.  
a. True  
*b. False  
46. Chlorine disinfection is more effective at higher pH  
a. True  
*b. False  
47. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the  
amount of residual chlorine remaining after a given contact time.  
a. True  
*b. False  
48. Chlorine feed rate and chlorine residual are both usually expressed in units of ppm.  
18.2 Chlorination  
281  
*a. True  
b. False  
49. Chlorine is used to sterilize wastewater in order to insure protection of public health.  
a. True  
*b. False  
50. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air  
a. True  
*b. False  
51. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air  
a. True  
*b. False  
52. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the  
amount of residual chlorine remaining after a given contact time.  
a. True  
*b. False  
53. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.  
a. True  
*b. False  
54. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports  
even when empty. Except when actually being moved to or from storage.  
*a. True  
b. False  
55. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.  
a. True  
*b. False  
56. Sodium hypochlorite is more effective and less expensive than chlorine gas  
a. True  
*b. False  
57. The chlorine dose rate plus the chlorine residual equals the chlorine demand.  
a. True  
*b. False  
58. Vents in a chlorine storage room are located at the ground level as chlorine is lighter than air  
a. True  
*b. False  
59. Chlorine demand is defined as the amount of chlorine remaining in the waste water at the end of a  
specific contact period.  
a. True  
*b. False  
60. Chlorine dosage is the difference between the amount of chlorine added to wastewater and the  
amount of residual chlorine remaining after a given contact time.  
a. True  
*b. False  
61. Hypochlorite solution is a more effective disinfectant than gas chlorine and is much less expensive.  
a. True  
*b. False  
282  
Chapter 18. Disinfection  
62. All 100- and 150-pound chlorine cylinders should be restrained or safety-chained to sturdy supports  
even when empty. Except when actually being moved to or from storage.  
*a. True  
b. False  
63. Chlorine demand is a good indicator of effective disinfection  
a. True  
*b. False  
64. A chlorine cylinder valve is thought to be leaking. If ammonia vapor is passed over the valve, the  
presence of a leak is indicated by  
a. A hissing noise.  
*b. A white cloud.  
c. An odor of hydrogen sulfide.  
d. Red smoke  
65. A chlorine residual is often maintained in a plant effluent:  
a. to keep the chlorinator working.  
b. for control of fluctuation of wastewater flow.  
c. for testing purposes.  
*d. to protect the bacteriological quality of the receiving water.  
e. None of the above.  
66. Acids should never be added to chlorine solutions as they  
*a. Cause chlorine gas to be released.  
b. Corrode or "eat away" the solution tank.  
c. Decrease the disinfecting properties of chlorine.  
d. Result in the formation of a chloride precipitate. „  
67. An amperometric titrater is used to measure  
a. Alkalinity.  
*b. Chlorine residual.  
c. Conductivity.  
d. COD.  
68. An operator should never enter a room containing a high concentration of chlorine gas without  
a. Staying low on the floor.  
b. Holding breath and have help standing by.  
*c. Having self-contained air or oxygen supply and help standing by.  
d. Covering nose and mouth with a wet handkerchief.  
69. As water temperatures decrease, the disinfecting action of chlorine  
*a. Decreases.  
b. Increases.  
c. Remains the same.  
70. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that  
is in service is normally determined by:  
a. knowing both the pressure and temperature of the cylinder pressure gauges.  
b. rotameter readings.  
*c. weighing of the cylinder or tank  
d. the chlorine residual test .  
e. None of the above.  
18.2 Chlorination  
283  
71. At what level should the exhaust be drawn off in a chlorination room?  
a. At chest height.  
b. At least two feet above the height of the chlorine cylinder.  
c. Near the ceiling.  
*d. Near the floor.  
72. Chloramines are  
*a. Combined chlorine.  
b. Enzymes.  
c. Found in polluted air.  
d. Free chlorine.  
73. Chlorine gas  
a. Is lighter than air.  
*b. Is heavier than air.  
c. Is pink in color.  
d. Will liquify at 70 degrees F.  
74. Chlorine gas is  
a. Colorless.  
*b. Heavier than air.  
c. Non-toxic.  
d. Odorless.  
75. Chlorine is:  
a. Colorless  
b. Explosive  
*c. Toxic  
d. All of the above  
76. Chlorine is being applied at a constant dose rate of 24 mg/L to a partially nitrified activated sludge  
effluent having a pH of 6.8 and a temperature of 67F. Ammonia-nitrogen is found to range from 2  
to 3 mg/L in this effluent. Disinfection in this effluent might be difficult because:  
a. a temperature of 70’ F or higher is necessary in order to achieve effective disinfection.  
b. chloramines are present most of the time.  
c. the chlorine dose rate is too low.  
d. the pH of this effluent will limit the effectiveness of free chlorine.  
*e. the ratio of chlorine to ammonia-nitrogen may make it difficult at times to maintain adequate  
chlorine residual.  
77. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
78. Chlorine is used to  
*a. Disinfect.  
b. Prevent corrosion.  
c. Raise the pH.  
d. Stabilize organics.  
284  
79. Chlorine residual may be determined using the reagent  
Chapter 18. Disinfection  
*a. Diethyl-p-phenylenediamine (DPD).  
b. Ethylendiamine tetraacetic acid (EDTA).  
c. Polychlorinated biphenyls (PCB).  
d. Sodium thiosulfate (Na2S203)  
80. "Chlorine residual" refers to:  
a. the amount of chlorine remaining in the ton cylinder after use.  
b. the amount of chlorine consumed during disinfection.  
*c. the chlorine remaining after disinfection.  
d. the chlorine that displays no disinfection power.  
e. the residue left after the evaporation of chlorine gas.  
81. The effectiveness of chlorine disinfection is measured by:  
a. the chlorine demand  
b. the chlorine dosage  
c. the total chlorine residual  
*d. the coliform concentration of the effluent  
82. The amount of chlorine used per day from a 1 ton chlorine cylinder is normally determined by:  
a. Pressure gauges.  
b. Rotometers.  
*c. Weighings.  
d. Chlorine residuals.  
e. Ammonia equivalents.  
83. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?  
a. 100  
b. 250  
*c. 460  
d. 490  
84. Which of the following discharges would in general, require the lowest chlorine dosage to ensure  
adequate disinfection?  
a. Primary plant effluent  
b. Activated sludge plant effluent  
c. Trickling filter plant effluent  
*d. Sand filter effluent  
e. Stabilization pond effluent  
85. Which of the following are factors that may influence the effectiveness of chlorine?  
a. Chlorine dose rate  
b. Contact time  
c. Suspended solids concentration of the wastewater being disinfected.  
*d. Only (a) and (b)  
e. (a), (b), and (c)  
86. The fundamental purpose of disinfection is to:  
a. Destroy fecal coliform bacteria  
b. Destroy all bacteria  
*c. Destroy pathogenic organisms  
d. Protect downstream users from waterborne diseases  
18.2 Chlorination  
285  
87. In the application of chlorine for disinfection, which of the following is not normally an operational  
consideration?  
a. Mixing  
b. Contact time  
*c. DO  
d. pH  
e. None of the above  
88. "Chlorine residual" refers to:  
a. the amount of chlorine remaining in the ton cylinder after use.  
b. the amount of chlorine consumed during disinfection.  
*c. the chlorine remaining after disinfection.  
d. the chlorine that displays no disinfection power.  
e. the residue left after the evaporation of chlorine gas.  
89. In the application of chlorine for disinfection, which of the following is not normally an operational  
consideration?  
a. Mixing  
b. Contact time  
*c. DO  
d. pH  
e. None of the above  
90. A chlorine residual is often maintained in a plant effluent:  
a. to keep the chlorinator working.  
b. for control of fluctuation of wastewater flow.  
c. for testing purposes.  
*d. to protect the bacteriological quality of the receiving water.  
e. None of the above.  
91. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that  
is in service is normally determined by:  
a. knowing both the pressure and temperature of the cylinder pressure gauges.  
b. rotameter readings.  
*c. weighing of the cylinder or tank  
d. the chlorine residual test .  
e. None of the above.  
92. The fundamental purpose of disinfection is to:  
a. Destroy fecal coliform bacteria  
b. Destroy all bacteria  
*c. Destroy pathogenic organisms  
d. Protect downstream users from waterborne diseases  
93. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?  
a. 100  
b. 250  
*c. 460  
d. 490  
94. Identify the incorrect statement regarding disinfection.  
a. When chlorine is added to water it forms acids, which tend to lower the pH of the wastewater  
286  
Chapter 18. Disinfection  
effluent  
b. HTH is a dry form of calcium hypochlorite  
c. Appropriate doses of chlorine may be used to control odors, control filamentous bulking in  
activated sludge mixed liquor, or reduce BOD5 of wastewater  
*d. Hypochlorite’s are sometimes used in place of chlorine because they are more effective and less  
costly  
95. Hypochlorite solution is used in effluent disinfection because:  
a. Chlorine residual determination is more stable and accurate in hypochlorite  
b. Hypochlorite residuals are more resistant to nitrite interference  
*c. Chlorine gas is more hazardous to store and handle  
d. Hypochlorite solution is easier and less costly to ship than gas chlorine  
e. Chlorine causes too many problems with disinfection efficiency  
96. Chlorine is:  
a. Colorless  
b. Explosive  
*c. Toxic  
d. All of the above  
97. The effectiveness of chlorine disinfection is measured by:  
a. the chlorine demand  
b. the chlorine dosage  
c. the total chlorine residual  
*d. the coliform concentration of the effluent  
98. The amount of chlorine used per day from a 1 ton chlorine cylinder is normally determined by:  
a. Pressure gauges.  
b. Rotometers.  
*c. Weighings.  
d. Chlorine residuals.  
e. Ammonia equivalents.  
99. Which of the following discharges would in general, require the lowest chlorine dosage to ensure  
adequate disinfection?  
a. Primary plant effluent  
b. Activated sludge plant effluent  
c. Trickling filter plant effluent  
*d. Sand filter effluent  
e. Stabilization pond effluent  
100. Which of the following are factors that may influence the effectiveness of chlorine?  
a. Chlorine dose rate  
b. Contact time  
c. Suspended solids concentration of the wastewater being disinfected.  
*d. Only (a) and (b)  
e. (a), (b), and (c)  
101. The fundamental purpose of disinfection is to:  
a. Destroy fecal coliform bacteria  
b. Destroy all bacteria  
*c. Destroy pathogenic organisms  
18.2 Chlorination  
d. Protect downstream users from waterborne diseases  
287  
102. In the application of chlorine for disinfection, which of the following is not normally an operational  
consideration?  
a. Mixing  
b. Contact time  
*c. DO  
d. pH  
e. None of the above  
103. "Chlorine residual" refers to:  
a. the amount of chlorine remaining in the ton cylinder after use.  
b. the amount of chlorine consumed during disinfection.  
*c. the chlorine remaining after disinfection.  
d. the chlorine that displays no disinfection power.  
e. the residue left after the evaporation of chlorine gas.  
104. In the application of chlorine for disinfection, which of the following is not normally an operational  
consideration?  
a. Mixing  
b. Contact time  
*c. DO  
d. pH  
e. None of the above  
105. A chlorine residual is often maintained in a plant effluent:  
a. to keep the chlorinator working.  
b. for control of fluctuation of wastewater flow.  
c. for testing purposes.  
*d. to protect the bacteriological quality of the receiving water.  
e. None of the above.  
106. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that  
is in service is normally determined by:  
a. knowing both the pressure and temperature of the cylinder pressure gauges.  
b. rotameter readings.  
*c. weighing of the cylinder or tank  
d. the chlorine residual test .  
e. None of the above.  
107. The fundamental purpose of disinfection is to:  
a. Destroy fecal coliform bacteria  
b. Destroy all bacteria  
*c. Destroy pathogenic organisms  
d. Protect downstream users from waterborne diseases  
108. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?  
a. 100  
b. 250  
*c. 460  
d. 490  
109. Identify the incorrect statement regarding disinfection.  
288  
Chapter 18. Disinfection  
a. When chlorine is added to water it forms acids, which tend to lower the pH of the wastewater  
effluent  
b. HTH is a dry form of calcium hypochlorite  
c. Appropriate doses of chlorine may be used to control odors, control filamentous bulking in  
activated sludge mixed liquor, or reduce BOD5 of wastewater  
*d. Hypochlorite’s are sometimes used in place of chlorine because they are more effective and less  
costly  
110. Hypochlorite solution is used in effluent disinfection because:  
a. Chlorine residual determination is more stable and accurate in hypochlorite  
b. Hypochlorite residuals are more resistant to nitrite interference  
*c. Chlorine gas is more hazardous to store and handle  
d. Hypochlorite solution is easier and less costly to ship than gas chlorine  
e. Chlorine causes too many problems with disinfection efficiency  
111. Chlorine is:  
a. Colorless  
b. Explosive  
*c. Toxic  
d. All of the above  
112. The difference between the amount of chlorine added and the amount of chlorine remaining after  
the contact period is referred to as:  
*a. the chlorine demand  
b. free chlorine residual  
c. total chlorine residual  
d. combined chlorine residual  
e. free available chlorine  
113. The ultimate measure of the effectiveness of chlorination in disinfection is:  
a. the measurement of chlorine dosage  
b. meeting the chlorine demand of the wastewater  
c. the establishment of a chlorine demand  
*d. effective reduction of the coliform count  
e. none of the above  
114. A chlorine cylinder valve is thought to be leaking. If ammonia vapor is passed over the valve, the  
presence of a leak is indicated by  
a. A hissing noise.  
*b. A white cloud.  
c. An odor of hydrogen sulfide.  
d. Red smoke  
115. A chlorine residual is often maintained in a plant effluent:  
a. to keep the chlorinator working.  
b. for control of fluctuation of wastewater flow.  
c. for testing purposes.  
*d. to protect the bacteriological quality of the receiving water.  
e. None of the above.  
116. Acids should never be added to chlorine solutions as they  
*a. Cause chlorine gas to be released.  
18.2 Chlorination  
b. Corrode or "eat away" the solution tank.  
289  
c. Decrease the disinfecting properties of chlorine.  
d. Result in the formation of a chloride precipitate. „  
117. An amperometric titrater is used to measure  
a. Alkalinity.  
*b. Chlorine residual.  
c. Conductivity.  
d. COD.  
118. An operator should never enter a room containing a high concentration of chlorine gas without  
a. Staying low on the floor.  
b. Holding breath and have help standing by.  
*c. Having self-contained air or oxygen supply and help standing by.  
d. Covering nose and mouth with a wet handkerchief.  
119. As water temperatures decrease, the disinfecting action of chlorine  
*a. Decreases.  
b. Increases.  
c. Remains the same.  
120. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that  
is in service is normally determined by:  
a. knowing both the pressure and temperature of the cylinder pressure gauges.  
b. rotameter readings.  
*c. weighing of the cylinder or tank  
d. the chlorine residual test .  
e. None of the above.  
121. At what level should the exhaust be drawn off in a chlorination room?  
a. At chest height.  
b. At least two feet above the height of the chlorine cylinder.  
c. Near the ceiling.  
*d. Near the floor.  
122. Chloramines are  
*a. Combined chlorine.  
b. Enzymes.  
c. Found in polluted air.  
d. Free chlorine.  
123. Chlorine gas  
a. Is lighter than air.  
*b. Is heavier than air.  
c. Is pink in color.  
d. Will liquify at 70 degrees F.  
124. Chlorine gas is  
a. Colorless.  
*b. Heavier than air.  
c. Non-toxic.  
d. Odorless.  
125. Chlorine is:  
290  
Chapter 18. Disinfection  
a. Colorless  
b. Explosive  
*c. Toxic  
d. All of the above  
126. Chlorine is being applied at a constant dose rate of 24 mg/L to a partially nitrified activated sludge  
effluent having a pH of 6.8 and a temperature of 67F. Ammonia-nitrogen is found to range from 2  
to 3 mg/L in this effluent. Disinfection in this effluent might be difficult because:  
a. a temperature of 70’ F or higher is necessary in order to achieve effective disinfection.  
b. chloramines are present most of the time.  
c. the chlorine dose rate is too low.  
d. the pH of this effluent will limit the effectiveness of free chlorine.  
*e. the ratio of chlorine to ammonia-nitrogen may make it difficult at times to maintain adequate  
chlorine residual.  
127. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
128. Chlorine is used to  
*a. Disinfect.  
b. Prevent corrosion.  
c. Raise the pH.  
d. Stabilize organics.  
129. Chlorine residual may be determined using the reagent  
*a. Diethyl-p-phenylenediamine (DPD).  
b. Ethylendiamine tetraacetic acid (EDTA).  
c. Polychlorinated biphenyls (PCB).  
d. Sodium thiosulfate (Na2S203)  
130. "Chlorine residual" refers to:  
a. the amount of chlorine remaining in the ton cylinder after use.  
b. the amount of chlorine consumed during disinfection.  
*c. the chlorine remaining after disinfection.  
d. the chlorine that displays no disinfection power.  
e. the residue left after the evaporation of chlorine gas.  
131. The effectiveness of chlorine disinfection is measured by:  
a. the chlorine demand  
b. the chlorine dosage  
c. the total chlorine residual  
*d. the coliform concentration of the effluent  
132. The amount of chlorine used per day from a 1 ton chlorine cylinder is normally determined by:  
a. Pressure gauges.  
b. Rotometers.  
*c. Weighings.  
d. Chlorine residuals.  
18.2 Chlorination  
e. Ammonia equivalents.  
291  
133. Which of the following discharges would in general, require the lowest chlorine dosage to ensure  
adequate disinfection?  
a. Primary plant effluent  
b. Activated sludge plant effluent  
c. Trickling filter plant effluent  
*d. Sand filter effluent  
e. Stabilization pond effluent  
134. Which of the following are factors that may influence the effectiveness of chlorine?  
a. Chlorine dose rate  
b. Contact time  
c. Suspended solids concentration of the wastewater being disinfected.  
*d. Only (a) and (b)  
e. (a), (b), and (c)  
135. The fundamental purpose of disinfection is to:  
a. Destroy fecal coliform bacteria  
b. Destroy all bacteria  
*c. Destroy pathogenic organisms  
d. Protect downstream users from waterborne diseases  
136. In the application of chlorine for disinfection, which of the following is not normally an operational  
consideration?  
a. Mixing  
b. Contact time  
*c. DO  
d. pH  
e. None of the above  
137. "Chlorine residual" refers to:  
a. the amount of chlorine remaining in the ton cylinder after use.  
b. the amount of chlorine consumed during disinfection.  
*c. the chlorine remaining after disinfection.  
d. the chlorine that displays no disinfection power.  
e. the residue left after the evaporation of chlorine gas.  
138. In the application of chlorine for disinfection, which of the following is not normally an operational  
consideration?  
a. Mixing  
b. Contact time  
*c. DO  
d. pH  
e. None of the above  
139. A chlorine residual is often maintained in a plant effluent:  
a. to keep the chlorinator working.  
b. for control of fluctuation of wastewater flow.  
c. for testing purposes.  
*d. to protect the bacteriological quality of the receiving water.  
e. None of the above.  
292  
Chapter 18. Disinfection  
140. At a wastewater treatment plant. The amount of chlorine used in a day from a cylinder or tank that  
is in service is normally determined by:  
a. knowing both the pressure and temperature of the cylinder pressure gauges.  
b. rotameter readings.  
*c. weighing of the cylinder or tank  
d. the chlorine residual test .  
e. None of the above.  
141. The fundamental purpose of disinfection is to:  
a. Destroy fecal coliform bacteria  
b. Destroy all bacteria  
*c. Destroy pathogenic organisms  
d. Protect downstream users from waterborne diseases  
142. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?  
a. 100  
b. 250  
*c. 460  
d. 490  
143. Identify the incorrect statement regarding disinfection.  
a. When chlorine is added to water it forms acids, which tend to lower the pH of the wastewater  
effluent  
b. HTH is a dry form of calcium hypochlorite  
c. Appropriate doses of chlorine may be used to control odors, control filamentous bulking in  
activated sludge mixed liquor, or reduce BOD5 of wastewater  
*d. Hypochlorite’s are sometimes used in place of chlorine because they are more effective and less  
costly  
144. Hypochlorite solution is used in effluent disinfection because:  
a. Chlorine residual determination is more stable and accurate in hypochlorite  
b. Hypochlorite residuals are more resistant to nitrite interference  
*c. Chlorine gas is more hazardous to store and handle  
d. Hypochlorite solution is easier and less costly to ship than gas chlorine  
e. Chlorine causes too many problems with disinfection efficiency  
145. Chlorine is:  
a. Colorless  
b. Explosive  
*c. Toxic  
d. All of the above  
146. Exhaust from a chlorinator room should be taken from  
a. Anywhere–the location is not important.  
*b. At floor level.  
c. Close to the entrance.  
d. In the ceiling.  
147. How many pounds of chlorine gas is necessary to treat 4,000,000 gallons of wastewater at a dosage  
of 2 mg/L?  
a. 61 lbs.  
b. 65 lbs.  
18.2 Chlorination  
293  
*c. 67 lbs.  
d. 69 lbs.  
148. How much chlorine is needed to provide 10 mg/L dosage for a flow of 2 MGD?  
a. 75 lbs.  
b. 83 lbs.  
c. 130 lbs.  
*d. 167 lbs.  
149. Hypochlorite solution is used in effluent disinfection because:  
a. Chlorine residual determination is more stable and accurate in hypochlorite  
b. Hypochlorite residuals are more resistant to nitrite interference  
*c. Chlorine gas is more hazardous to store and handle  
d. Hypochlorite solution is easier and less costly to ship than gas chlorine  
e. Chlorine causes too many problems with disinfection efficiency  
150. Identify the incorrect statement regarding disinfection.  
a. When chlorine is added to water it forms acids, which tend to lower the pH of the wastewater  
effluent  
b. HTH is a dry form of calcium hypochlorite  
c. Appropriate doses of chlorine may be used to control odors, control filamentous bulking in  
activated sludge mixed liquor, or reduce BOD5 of wastewater  
*d. Hypochlorite’s are sometimes used in place of chlorine because they are more effective and less  
costly  
151. If a chlorinator is connected to the bottom valve of a half full one-ton cylinder, the chlorinator is  
withdrawing  
a. Chlorine gas.  
*b. Liquid chlorine.  
c. Liquid or gas chlorine, depending upon temperature.  
d. Nothing, since there is only one connection.  
152. If you encounter a liquid chlorine leak in a one-ton cylinder, what action will immediately help to  
reduce the effect of the leak?  
a. Spray it with water.  
b. Apply ice. to leaking area.  
c. Spray it with ammonia solution.  
*d. Rotate cylinder so leak is in uppermost position.  
153. In the application of chlorine for disinfection, which of the following is not normally an operational  
consideration?  
a. Mixing  
b. Contact time  
*c. DO  
d. pH  
e. None of the above  
154. One liter of liquid chlorine can evaporate and produce how many liters of chlorine gas?  
a. 100  
b. 250  
*c. 460  
d. 490  
294  
Chapter 18. Disinfection  
155. The amount of chlorine used per day from a 1 ton chlorine cylinder is normally determined by:  
a. Pressure gauges.  
b. Rotometers.  
*c. Weighings.  
d. Chlorine residuals.  
e. Ammonia equivalents.  
156. The amperometric titration method is used to measure:  
a. Alkalinity.  
*b. Chlorine residual.  
c. pH.  
d. Total hardness.  
157. The fundamental purpose of disinfection is to:  
a. Destroy fecal coliform bacteria  
b. Destroy all bacteria  
*c. Destroy pathogenic organisms  
d. Protect downstream users from waterborne diseases  
158. The measure of the effectiveness of chlorine in disinfection is:  
a. The chlorine demand.  
b. The chlorine dosage.  
*c. The chlorine residual.  
d. The amount of chloramine formed.  
e. The final effluent coliform concentration.  
159. When replacing piping in your chlorinator, be sure to  
a. Replace with steel lines only.  
b. Replace with CIP only.  
*c. Replace with corrosion resistant piping only.  
d. No special reqirement.  
160. Which of the following discharges would, in general, require the lowest chlorine dosage to ensure  
adequate disinfection?  
a. Primary plant effluent  
b. Activated sludge plant effluent  
c. Trickling filter plant effluent  
*d. Sand filter effluent  
e. Stabilization pond effluent  
161. Which of the following are factors that may influence the effectiveness of chlorine?  
a. Chlorine dose rate  
b. Contact time  
c. Suspended solids concentration of the wastewater being disinfected.  
*d. Only (a) and (b)  
e. (a), (b), and (c)  
162. Which of the following are safe procedures for handling chlorine cylinders?  
a. Keep cylinders close to direct heat to prevent freezing. -  
*b. Protective cap should always be in place when moving a cylinder.  
c. Roll cylinders in a horizontal position.  
d. Store cylinders on their sides.  
18.2 Chlorination  
295  
163. Which of the following discharges would in general, require the lowest chlorine dosage to ensure  
adequate disinfection?  
a. Primary plant effluent  
b. Activated sludge plant effluent  
c. Trickling filter plant effluent  
*d. Sand filter effluent  
e. Stabilization pond effluent  
19. Odor Control  
19.1 Background  
Odors associated with wastewater, result from the release of the following:  
1. compounds which are originally discharged into the sewer – chemicals, wastes, or  
2. by products of biological and chemical reactions occurring in wastewater  
Odors are generated from every phase of wastewater treatment. Odor characteristics, intensity, and volume  
of foul air generated is dictated by the treatment process and the stage of wastewater treatment.  
Odors are dependent on:  
• Odor causing compounds present  
• Exposed process surface area  
External factors including temperature and atmospheric conditions including wind speed and  
direction.  
• Proximity and sensitivity of odor receptors  
Key wastewater properties which dictate odors include:  
pH and Temperature: These affect the rate of biological degradation and the rate at which the  
odorous compounds are released.  
ORP: ORP governs the rate of biodegradation activities in the wastewater. Lower (more negative)  
implies anaerobic conditions.  
• Dissolved/bound/available oxygen: Presence of oxygen will increase the wastewater ORP.  
Conveyance time: Longer time due to longer distance and lower velocities will promote anaerobic  
degradation in the wastewater.  
19.2 Drivers for Odor Control  
1. Air Quality Regulations  
 
 
 
298  
Chapter 19. Odor Control  
In many instances, the need for odor control systems is driven by local and state air quality regula-  
tions including regulations related to prevent public nuisance.  
2. Low Odor Thresholds  
Many of the compounds commonly found in wastewater such as H2S, skatoles have perceptible  
odors even at very low concentrations at parts per billion levels (ppb).  
3. Worker Safety  
Ventilation requirements to protect workers form health hazard associated with compounds such  
as H2S and to comply with fire and explosion prevention related regulations including NFPA, will  
generate foul air requiring treatment prior to discharge.  
4. Corrosion Prevention Severe corrosion potential of wastewater conveyance and treatment infrastruc-  
ture due to H2S conversion by bacteria to sulfuric acid.  
5. Good Neighbor Policy The location of the treatment plants and sewage conveyance systems in  
urban areas necessitate treatment plants to adopt a good neighbor policy to establish expecta-  
tions/limits of the treatment plant vis-a-vis odor, noise, light pollution etc. to ensure harmony  
with its neighbors.  
19.3 Odor Causing Compounds  
Hydrogen sulfide(H2S) is the most common compound which causes odors related to wastewater treat-  
ment.  
• Hydrogen sulfide is the most common wastewater origin odorous compound  
it is produced in wastewater by the activity of sulfate reducing bacteria which live in the slime layer  
in the sewer pipes.  
• hydrogen sulfide characteristics:  
has an offensive smell  
it is highly toxic and has the potential to instantly kill  
it is converted into highly corrosive sulfuric acid through microbiological activity in the  
wastewater systems.  
Thus, the prevention or treatment of odors, in particular hydrogen sulfide provides benefits includ-  
ing:  
safety  
preventing public nuisance, and  
corrosion prevention.  
• Other common odor pollutants include  
organic compounds - these are typically associated with the foul air from the preliminary,  
primary and secondary treatment processes  
reduced sulfur compounds including mercaptans which are typically byproducts of solids  
decomposition  
ammonia - the odor causing constituent of the foul air from dewatering operations associated  
with anaerobic digested sludge.  
19.4 Theory of Odor Control  
Theory related to the control of the the main odor causing compounds is as follows:  
 
 
19.4 Theory of Odor Control  
299  
300  
Chapter 19. Odor Control  
19.4.1 Hydrogen Sulfide  
H2S is generated in the wastewater from the biological reduction of dissolved sulfates and some of the  
H2S present in the wastewater will escape into the gas phase causing odors. Once formed, H2S control is  
accomplished using the following principles:  
1. pH Control  
• Amount of H2S escaping into the gas phase is dictated by Henry’s Law and is pH dependent.  
H2S being a weak acid, when present in an alkaline (>7 pH) solution, will ionize to HS−  
(bisulfide) and subsequently to S-(sulfide) ions.  
• H2S odors will not occur as long as H2S remains in solution as HSor S−  
Increasing pH (or OHconc.) does not destroy H2S, but keeps it from escaping into the gas  
phase, as long as an alkaline pH is maintained.  
By adding alkaline chemicals, a majority of H2S remains in solution in the wastewater and  
the amount of odorous H2S released in the gas phase is reduced.  
2. Chemical Precipitation  
H2S can be precipitated using iron salts such as ferric or ferrous chloride.  
3. Chemical Oxidation  
Strong oxidants including bleach and hydrogen sulfide can oxidize H2S to elemental sulfur.  
19.4.2 Ammonia  
Ammonia (NH3) is produced from the bio degradation of nitrogenous material in wastewater  
including proteins and uric acid.  
NH3 is soluble in water and when in solution (liquid phase), some NH3 will escape into the gas  
phase causing odors.  
Amount of NH3 in the gas phase is dictated by Henry’s Law and is pH dependent NH3 is a weak  
base (unlike H2S which is a weak acid) and in the presence of H+, it ionizes to NH4+ (ammonium).  
Under acidic conditions, NH3 is kept in the liquid phase of wastewater, reducing amount of odor-  
ous NH3 released in the gas phase.  
NH3 odors will not occur as long as NH3 remain in solution as NH4+. pH adjustment only helps to  
keep the ammonia in solution, it does not remove/destroy the ammonia.  
19.5 Odor Control Technologies  
The odor control technologies used in wastewater treatment can be classified into two categories:  
1. Liquid Phase Odor Control  
2. Vapor Phase Odor Control  
19.6 Liquid Phase Odor Control Methods  
• typically applied in the collections systems  
goal for this treatment is to prevent nuisance odors associated with the production and release of  
hydrogen sulfide  
Liquid-phase odor control strategies include:  
1. Chemical precipitation:  
Iron salts react with the hydrogen sulfide present in the wastewater to form insoluble iron sulfide  
precipitate  
 
 
 
 
19.6 Liquid Phase Odor Control Methods  
301  
(a) H2S - H2SpH Equilibrium Curve  
(b) NH3 - NH+4 pH Equilibrium Curve  
2. Nitrate addition:  
Addition of nitrate salts promotes the activity of certain bacteria such as Thiobacillus denitrificans,  
typically present in wastewater, which oxidize reduced sulfur compounds (like H2S) while den-  
302  
Chapter 19. Odor Control  
19.6 Liquid Phase Odor Control Methods  
303  
itrifying the nitrate. Presence of nitrate, also, increases oxidation-reduction potential, inhibiting  
the production of any odorous compounds such as hydrogen sulfide which are produced under  
anaerobic conditions.  
3. pH Control:  
Addition of alkaline chemicals including caustic soda or magnesium hydroxide increases the pH  
of the wastewater, preventing the escape of the hydrogen sulfide to the air phase as under alkaline  
condition H2S is present as HS.  
By raising the pH of wastewater through the addition of alkaline chemicals—caustic soda  
(NaOH) or magnesium hydroxide (Mg(OH)2), minimizes the potential of releasing odorous  
H2S to the gas phase.  
This treatment does not destroy the H2S but reduces its potential to return back to the gas  
phase as long as the alkaline pH is maintained. Caustic soda also helps remove the slime  
layer in the collections piping where the anaerobic bacteria responsible for the formation of  
hydrogen sulfide are present.  
• Magnesium hydroxide Mg(OH)2 can also be used for raising the pH.  
• Mg(OH)2 is used primarily because of it being environmentally safer compared to NaOH  
Mg(OH)2 is supplied as specially formulated slurry to improve solubility, keeping the solids  
in suspension and prevent solids from settling and getting deposited.  
Advantage of using Mg(OH)2  
Poses less environmental risk than NaOH due to an accidental release because of its lower pH  
and corrosivity.  
304  
Chapter 19. Odor Control  
Figure 19.4: Speece Cone  
Disadvantages of using Mg(OH)2  
• Lower effectiveness range  
• Not effective for higher H2S concentrations  
• Higher cost  
4. Air/oxygen injection The injection of air or oxygen in the sewer conveyance systems prevents  
developing anaerobic condition and thus formation of hydrogen sulfide. A Speece Cone allows for  
the injection of pure oxygen into the sewage.  
19.7 Vapor Phase Odor Control Methods  
• is typically applied inside the plant  
foul air from the treatment processes is captured and scrubbed using either a packed tower scrubber,  
a carbon scrubber or a biofilter.  
19.7.1 Common Vapor Phase Odor Control Methods  
1. Packed Tower Scrubbers Packed tower scrubbers are usually rectangular or cylindrical shaped.  
• Foul air is injected from the bottom  
• Chemicals are added to the recirculating water  
• Recirculated liquor from the sump is sprayed from the top  
Plastic packing media provides the surface to facilitate the transfer of pollutants from the gas  
phase to the liquid phase.  
The pollutants transferred to the liquid phase are chemically or biologically removed or  
stabilized so it does not return back to the gas phase.  
Recirculation water is wasted periodically by adding make-up water to prevent build-up of  
the pollutant in the recirculation liquor.  
• Demister prevents the carryover of water with the air leaving the scrubber at top.  
 
 
19.7 Vapor Phase Odor Control Methods  
305  
Figure 19.5: Packed Tower Scrubber  
• Chemical oxidants such as hydrogen peroxide (H2O2) and bleach (NaOCl), added to the recir-  
culation water in the packed tower scrubber, removes the odor causing pollutants chemically.  
Soluble organics are oxidized to less odorous compounds and reduced sulfur compounds are  
converted to elemental sulfur.  
Solubility is the key! Only those pollutants which are soluble in water and transferred from  
the gas phase to liquid phase in the packed tower scrubber will be oxidized.  
306  
Chapter 19. Odor Control  
Applications of Packed Tower Scrubber  
For controlling organics using oxidants:  
An oxidant such as hydrogen peroxide or bleach is added to recirculation water.  
Organics control is typically exercised for foul air from the preliminary and secondary  
treatment processes.  
Only organics soluble in water will be removed  
For controlling hydrogen sulfide using chemicals:  
Hydrogen sulfide control in a packed tower scrubber using chemicals is accomplished us-  
ing either oxidants such as hydrogen peroxide and bleach (which is a solution of sodium  
hypoclorite in caustic soda) or alkaline chemicals such as caustic soda or bleach can be added  
to recirculation water. Note: Bleach is both - oxidant and alkaline.  
H2S transferred from the gas phase to the liquid phase in the packed tower scrubber is  
kept in the liquid phase by addition of alkaline chemicals—typically bleach (NaOCl) or  
caustic soda (NaOH).  
This treatment does not destroy the H2S but reduces its potential to return back to the  
gas phase.  
Bleach solution used is typically supplied as a solution contains 12.5% active chlorine  
and has a pH of about 11.5.  
Caustic soda has a pH of 14  
Bleach is advantageous over caustic for the following reasons:  
Use of a lower pH bleach reduces the hardness precipitation potential thereby  
leading to less media blockage and therefore less acid washing need.  
Oxidizes odorous organics  
*
*
For controlling ammonia using chemicals:  
NH3 is produced from the bio degradation of nitrogenous material in wastewater includ-  
ing proteins and uric acid.  
NH3 is soluble in water and when in solution (liquid phase), some NH3 will escape into  
the gas phase causing odors.  
Amount of NH3 in the gas phase is dictated by Henry’s Law and is pH dependent NH3  
is a weak base (unlike H2S which is a weak acid) and in the presence of H+, it ionizes to  
NH+4 (ammonium). Under acidic conditions, NH3 is kept in the liquid phase of wastewa-  
ter, reducing amount of odorous NH3 released in the gas phase.  
NH3 odors will not occur as long as NH3 remain in solution as NH4+. pH adjustment  
only helps to keep the ammonia in solution, it does not remove/destroy the ammonia.  
An acid like sulfuric acid, can be used for lowering the pH of the packed tower recircula-  
tion water to convert NH3 to NH4+.  
In typical wastewater odor control applications, the foul air ammonia concentration does  
not warrant use of an acid.  
Solubility of ammonia in water itself is sufficient to provide the control of ammonia.  
As a biotrickling filter for controlling H2S and organics:  
Packed tower scrubbers can be operated as bioscrubbers, also known as biotrickling  
filters.  
Bacteria in the slime layer on the packing and in the recirculation water consume the  
pollutants.  
Aerobic process promotes growth of sulfur oxidizing bacteria when treating foul air  
19.7 Vapor Phase Odor Control Methods  
307  
with H2S  
When used for controlling H2S, the pH of the recirculation water drops because of the  
biological conversion of H2S to sulfuric acid.  
308  
2. Carbon Scrubbers  
Chapter 19. Odor Control  
This odor scrubber utilizes activated carbon’s natural adsorptive property wherein odor caus-  
ing compounds are attracted and held to its surface.  
The carbon adsorbs pollutants from the foul air passing through the scrubber. Carbon scrub-  
bers can be designed to remove specific target pollutants.  
• Suitable for non-polar organic and inorganic compounds including H2S  
The carbon may be impregnated with an oxidant such as potassium permanganate or a al-  
kaline substrate to enhance its effectiveness in treating foul air with organics and hydrogen  
sulfide respectively..  
When the capacity of the carbon bed is exhausted, the carbon can be cleaned/regenerated or  
replaced.  
Figure 19.6: Carbon Scrubber  
19.7 Vapor Phase Odor Control Methods  
309  
Figure 19.7: Biofilter  
3. Biofilters  
Biofilters utilize biological processes wherein microorganisms growing on a slime layer on  
a packed substrate - biofilter media, consume the odor causing compounds from the foul air  
which dissolve in the slime layer as it passes through the biofilter  
• Biofilters can remove hydrogen sulfide, organics and ammonia  
• Typical biofilter media includes organic or inert material such as compost or activated carbon.  
• Foul air is passed through the bed from the bottom  
Irrigation water is provided on the surface and/or the foul air stream is humidified to sustain  
the biological growth  
Advantages of biofilters include:  
• Simplicity of design and operation  
• Minimal maintenance requirements  
310  
4. Ozonation  
Chapter 19. Odor Control  
Ozone, a strong oxidizing agent, is injected in conjunction with water in the headspace to  
control odors.  
Application involves the use of an on-site ozone generator for controlling odors at locations  
such as a pump station wet well.  
• Effective method to control odors in sensitive environments near residential neighborhoods.  
The ozone based oxidation of compounds including H2S and organics requires a very short  
contact time.  
Figure 19.8: Ozone Injection Unit  
Chapter Assessment  
1. Three systems used for vapor phase odor control in wastewater treatment  
2. Explain how bioxide works to control H2S odor  
3. Why do we need to control H2S in wastewater/wastewater treatment  
4. Explain how bioxide (sodium nitrate) works in controlling odors  
5. In your own words how is a biofilter different from a carbon scrubber  
6. Why is the control of hydrogen sulfide important in wastewater treatment  
7. Explain how bioxide works to control H2S odor  
8. Why do we need to control H2S in wastewater/wastewater treatment  
9. What are the characteristics of hydrogen sulfide  
10. Explain how bioxide (sodium nitrate) works in controlling odors  
11. In your own words how is a biofilter different from a carbon scrubber  
12. Why is the control of hydrogen sulfide important in wastewater treatment  
13. Three systems used for vapor phase odor control in wastewater treatment  
14. Hydrogen sulfide removal from the foul air can be accomplished by scrubbing with an alkaline  
solution  
*a. True  
b. False  
15. Odor control of hydrogen sulfide can be accomplished by the use of which of the following agents?  
a. hydrogen peroxide  
b. chlorine  
c. ozone  
*d. all of the above  
e. none of the above  
16. Ferric chloride helps in odor control by:  
a. Oxidizing the odor constituents  
b. Destruction of microorganisms responsible for odors  
312  
Chapter 19. Odor Control  
*c. Precipitating hydrogen sulfide  
d. Raising the pH of the wastewater  
17. Use of caustic soda in odor scrubbers is used for controlling:  
*a. Hydrogen sulfide  
b. Ammonia  
c. Fouling  
d. Organic compounds  
18. Caustic soda is used in odor scrubbers for controlling:  
*a. Hydrogen sulfide  
b. Ammonia  
c. Fouling  
d. Organic compounds  
19. Hydrogen sulfide control in the collection systems by caustic soda dosing is accomplished by:  
*a. pH control  
b. Chemical reaction  
c. Oxidation  
d. Biological control  
20. High sulfide concentrations (either gaseous or dissolved) often cause problems in a wastewater  
treatment plant’s influent structure. Pre-chlorination of a plant influent is routinely practiced to  
control sulfides and odors. At your plant the piping that supplies chlorine for pre-chlorination needs  
replacing. It is developing cracks and is corroded. Rather than replace this piping immediately, you  
are directed by your supervisor to identify alternate means of controlling these influent sulfides. Do  
the following:  
(a) Identify THREE alternative methods of controlling sulfides.  
(b) Briefly explain how each of these methods identified in (1) is able to control sulfides (e.g  
chlorine destroys sulfides by chemical oxidization of these sulfides.  
(c) From the three alternative methods of controlling sulfides select the one you feel is the best  
substitute for chlorine gas and briefly state why you think it is best.  
Response:  
a Identify THREE alternative methods of controlling sulfides.  
• pH control – Caustic/magnesium hydroxide addition in the collection system.  
• Foul air treatment using chemical scrubbers – capture and treat the foul air  
• Chemical precipitation using iron salts  
b
Briefly explain how each of these methods identified in (1) is able to control sulfides (e.g  
chlorine destroys sulfides by chemical oxidization of these sulfides.  
pH control: Keeps the H2S in the liquid phase. Caustic would also help in control-  
ling/removing the slime layer which is responsible for odor/H2S generation.  
Foul air treatment using chemical or biological scrubbers – capture and treat the foul air  
– for chemical scrubbers - methods used could be using an oxidizing agent (peroxide or  
beach) or alkaline pH (using caustic) recirculation water.  
Chemical precipitation using iron salts – iron salts chemically remove the hydrogen  
sulfide by forming iron sulfide precipitate.  
c
From the three alternative methods of controlling sulfides select the one you feel is the best  
substitute for chlorine gas and briefly state why you think it is best.  
Process changes to remove ammonia further through implementation of nitrification –  
19.7 Vapor Phase Odor Control Methods  
313  
denitrification as part of the activated sludge process  
• Iron sulfide – good control over anticipated sulfide levels in the plant  
21. Why is H2S control important? Describe the methods for controlling odor in both, in the sewer  
collection systems and in the plant.  
Response:  
a Why is H2S control important?  
• safety  
• preventing public nuisance, and  
• corrosion prevention.  
b
Describe the methods for controlling odor in both, in the sewer collection systems and in the  
plant.  
In sewer collection system:  
i. pH adjustment by dosing with alkaline chemicals such as caustic soda or magnesium  
hydroxide  
ii. By injecting oxygen or air to prevent the formation of H2S  
iii. By dosing with a iron salt to precipitate the H2S  
iv. By adding sodium nitrate (Bioxide) to facilitate biodegradation of H2S  
In the plant:  
By capturing and treating the foul air using the following methods:  
i. Packed tower chemical scrubber using one of the following:  
• Oxidizing chemicals such as hydrogen peroxide or bleach  
• Chemicals like caustic soda which increase the pH of the recirculation water  
ii. Use of biological removal based systems such as:  
• Biofilter  
• Biotrickling filter  
iii. Use of adsorbtive material such as activated carbon  
22. What is UV radiation disinfection? Discuss maintenance and operational issues. Advantages and  
disadvantages. What is the dosage measured in and what does the equation (MW-sec/cm2) mean?  
 
20. Wastewater Chemicals  
20.1 Wastewater treatment chemicals - by use/category  
USE/CATEGORY  
pH Control/  
Alkalinity Supplement  
PROCESS  
CHEMICALS USED  
Caustic soda  
Odor Control  
Secondary Treatment  
Digestion  
Magnesium hydroxide  
Calcium oxide  
Ammonia  
Sodium carbonate  
Muriatic acid  
Oxidant  
Odor Control  
Disinfection  
Chlorine  
Sodium hypochlorite (NaOCl)  
Calcium hypochlorite (HTH)  
Hydrogen Peroxide  
Advance Primary Treatment/  
Chemically Enhanced Primary Treat-  
ment (CEPT)/Phys-Chem  
Primary Treatment  
Secondary  
Ferric Chloride  
Anionic Polymer  
Filament Control  
Bleach  
Cationic Polymer  
Phosphorous Removal  
Nitrogen Removal  
Dechlorination  
Primary Treatment  
Iron Salts  
Secondary Treatment  
Alum (Precipitant)  
Breakpoint Chlorination  
Chlrorine  
Sodium Hypochlorite  
Disinfection  
Sodium bisulfite  
Sulfur dioxide  
Flocculation/Solids Separation  
Descaling  
Sludge Dewatering  
Cationic Polymer  
Odor Control Scrubber  
Muriatic Acid  
 
20.2 Wastewater treatment chemicals - by use/category  
317  
20.2 Wastewater treatment chemicals - by use/category  
PROCESS  
Collections  
ACTION  
Odor Control  
CHEMICAL USED (ROLE)  
Caustic Soda (pH control)  
Magnesium Hydroxide (pH control)  
Hydrogen Peroxide (Oxidant)  
Sodium Nitrate (Biological Degradation)  
Iron Salts (Precipitant)  
Primary  
CEPT  
Ferric Chloride (Coagulant)  
Anionic Polymer (Flocculant)  
Secondary  
Filament Control  
Bleach  
Cationic Polymer  
Cationic Polymer (Flocculant)  
WAS Thickening  
Nutrient Removal  
Tertiary Treatment  
Phosphorous Removal  
Iron Salts (Precipitant)  
Alum (Precipitant)  
Calcium Oxide  
Alkalinity Supplementation  
Ammonia  
Sodium Carbonate  
Disinfection  
Chlorine/Bleach  
Sodium Bisulfite  
Sulfur Dioxide  
Dechlorination  
Dewatering  
Flocculation  
Cationic Polymer  
Plant Odor Control  
Foul Air Scrubbing  
Hydrogen Peroxide (Oxidant)  
Bleach (Oxidant)  
Caustic Soda (pH Control)  
Muriatic Acid (pH Control & Scrubber  
Descaling)  
Anaerobic Digestion  
Hydrogen Sulfide Control  
Alkalinity Supplementation  
Iron Salts (Precipitant)  
Calcium Oxide  
Ammonia  
Sodium Carbonate  
20.2.1 Polymers in wastewater treatment  
• Polymer use in wastewater treatment includes:  
For enhancing primary removal efficiencies  
For sludge thickening - to increase the solids content of the sludge feed to the digester  
For solids dewatering - to reduce the digested solids hauling cost and to make the final solids  
product more manageable  
For filament control in activated sludge treatment  
Both anionic and cationic polymers used in wastewater treatment are available in the following forms:  
1. Dry Polymers: These are available in granular, flake or bead form and have an active polymer as  
high as 95%. Prior to use, the dry polymers have to be dissolved in water using specialized mixing  
 
 
318  
Chapter 20. Wastewater Chemicals  
units  
2. Emulsion Polymer: This water soluble version consists of water droplets dispersed in oil. They  
have 25% to 50% active polymer content and require a specialized system to disperse it in water  
prior to use.  
3. Solution polymers: These are water soluble polymers in water. These polymers are relatively easy  
to put into dilute solution. However, the lower active polymer content increases the shipping cost of  
this type of polymer.  
4. Cationic polymer is also available as a low cost solution type polymer - Mannich Polymer (man-  
nich is a type of chemical reaction involving formalydehyde which is used for making this poly-  
mer). However, it has certain drawbacks which include:  
(a) Presence of formaldehyde which lends its offensive odor  
(b) Higher viscosity which imposes operational challenges related to its use, and  
(c) High pH which leads to formation of hardness deposits in the associated piping and equip-  
ment.  
The polymers use is primarily a function of the process stream. Each system is different and there are no  
hard and fast rules regarding which products will work and therefore jar tests and pilot tests are conducted  
as part of the product selection process.  
20.3 Chemical dosing math problems  
20.3.1 lbs chemicals needed given flow and dosing rate  
• Use lbs formula to calculate the lbs of chemicals required  
Using the calculated lbs chemical required value, calculate the amount of that chemical at the  
concentration available  
So for example, if asked how much many gallons per day of bleach solution (SG 1.2)containing 12.5%  
available chlorine is required to disinfect a 10 MGD flow of water given the required chlorine dosage of 7  
mg/l.  
1. calculate the lbs of chlorine required using the lbs formula:  
mg  
l
=10MGD 7  
8.34 = 583.8 lbs chlorine per day  
2. calculate the gallons of bleach which will provide the 583.3 lbs chlorine  
Applying the lbs formula - note that 8.34 * SG will give the actual lbs/gal of bleach. If SG is not  
provided, use only 8.34 lbs per gallon:  
lbs bleach  
day  
gal  
day  
lbs bleach lbs chlorine  
0.0125  
gal lb bleach  
583.3  
= x  
8.341.2  
gal  
day  
583.3  
8.341.20.125  
gal  
day  
=x  
=
= 466  
20.3.2 Chemical batching and dilution  
These problems include questions such as: How much initial volume of a 4% polymer solution is needed  
to make 3500 gallons of polymer at 0.25% concentration?  
 
 
 
20.3 Chemical dosing math problems  
319  
These type of problems are solved using C*V relationship where C is the concentration and V is  
the volume.  
As C is expressed in weight/volume, C*V will equal to weight. The weight of the chemical will be  
same before and after the dilution  
If C1 is the concentration of the chemical before dilution and V1 is the volume of that initial con-  
centration that is needed and C2 is the final concentration that you want to make and V2 is the  
volume that you are making of the final concentration, C1 * V1 = C2 * V2.  
• Knowing C1, C2 and V2, we can calculate V1 as:  
C2 V2  
V1 =  
C1  
C.25% V.25%  
0.25 3500  
V4%  
=
=
= 219gal  
C4%  
4
Take 219 gallons of the 4% polymer and dilute to 3,500 gallons to give a 0.25% polymer solution.  
Chapter Assessment  
1. What is mannich polymer and what are its drawbacks.  
2. Chemical requirements for the conditioning of sludge are normally based upon laboratory-scale "jar  
tests" which determine the volume of chemical solution required for floc formation.  
*a. True  
b. False  
3. Cationic polymers are high-molecular-weight organic compounds carrying a negative charge  
a. True  
*b. False  
4. Anionic polymer is used for:  
a. Thickening solids in a gravity thickener  
b. Flocculating solids for dewatering  
c. For odor control  
*d. For enhancing solids and BOD removal in the primaries  
5. Alum is frequently used along with an anionic polymer when dewatering anaerobically digested  
sludge using a belt press.  
a. Cationic polymers are high molecular weight organic compounds carrying a negative charge.  
b. A dry polymer is always a better choice for application in centrifuges than any liquid polymer  
solution.  
*c. Because of its viscosity, a Mannich polymer may be difficult to pump.  
d. All liquid polymer solutions are harmless and need not require the examination of their MSDS.  
6. Either alum; ferric chloride; or lime may be used to remove solids from a secondary effluent.  
Which one of, the following statements is TRUE regarding these chemicals:  
a. Typical dose rates for alum when it is applied for the removal of phosphorus from a secondary  
effluent are 1 to 10 mg/L.  
b. Hydrated lime needs to be "slaked" prior to use.  
*c. The safety precautions for handling liquid ferric chloride are the same as those for handling an  
322  
Chapter 20. Wastewater Chemicals  
acid.  
d. All of these chemicals raise the pH of the wastewater to which they are applied.  
7. Ferric chloride helps in odor control by:  
a. Oxidizing the odor constituents  
b. Destruction of microorganisms responsible for odors  
*c. Precipitating hydrogen sulfide  
d. Raising the pH of the wastewater  
8. Flocculation is best accomplished by  
a. Decreasing alkalinity.  
*b. Gentle agitation.  
c. Increased sunlight.  
9. Sodium hydroxide, (caustic soda) when used in wastewater:  
Is typically applied at 1 - 10 mg/l when used to precipitate phosphorus in primary sedimentation  
systems.  
a. Should be treated as an acid with regard to safe handling.  
b. Should be immediately diluted to 10% upon receiving.  
*c. Raises the pH of the wastewater to which it is added.  
d. Is added to filtered effluent to improve de-chlorination with sulfur dioxide.  
10. Either alum; ferric chloride; or lime may be used to remove solids from a secondary effluent.  
Which one of, the following statements is TRUE regarding these chemicals:  
a. Typical dose rates for alum when it is applied for the removal of phosphorus from a secondary  
effluent are 1 to 10 mg/L.  
b. Hydrated lime needs to be "slaked" prior to use.  
*c. The safety precautions for handling liquid ferric chloride are the same as those for handling an  
acid.  
d. All of these chemicals raise the pH of the wastewater to which they are applied.  
11. Flocculation is best accomplished by  
a. Decreasing alkalinity.  
*b. Gentle agitation.  
c. Increased sunlight.  
d. Rapid mixing  
12. Flocculation is best accomplished by:  
a. Decreasing alkalinity  
*b. Gentle agitation  
c. Increased sunlight.  
13. If a chemical costs S30 per ton, how much will it cost per year to treat a flow of 1.5 MGD if the  
average dose is 18 mg/L?  
a. $803.  
b. $110.  
*c. $233.  
d. $506.  
Identify the correct statement regarding polymers.  
14. Sodium hydroxide, (caustic soda) when used in wastewater:  
Is typically applied at 1 - 10 mg/l when used to precipitate phosphorus in primary sedimentation  
systems.  
20.3 Chemical dosing math problems  
323  
a. Should be treated as an acid with regard to safe handling.  
b. Should be immediately diluted to 10% upon receiving.  
*c. Raises the pH of the wastewater to which it is added.  
d. Is added to filtered effluent to improve de-chlorination with sulfur dioxide.  
15. Which one of the following statement is TRUE regarding polymers?  
a. Alum is frequently used along with an anionic polymer when dewatering anaerobically digested  
sludge using a belt press.  
b. Cationic polymers are high-molecular-weight organic compound carrying a negative charge.  
c. A dry polymer is always a better choice for application in centrifuges than any liquid polymer  
solution.  
*d. Because of its high pH, Mannich polymers may cause scale formation.  
e. All. liquid polymer solutions are harmless and need not require the examination of its MSDS  
sheet.  
16. Which one of the following statement is TRUE regarding polymers?  
a. Cationic polymers are high molecular weight organic compound carrying a negative charge.  
b. A dry polymer is always a better choice for application in centrifuges than any liquid polymer  
solution.  
*c. Because of its high pH, Mannich polymers may cause scale formation.  
d. All liquid polymer solutions are harmless and need not require the examination of its MSDS  
sheet.  
e. Alum is frequently used along with an anionic polymer when dewatering anaerobically digested  
sludge using a belt press.  
21. Pumping Efficiencies and Costs  
Pump is a machine used for moving water (and other fluids) through a piping system and raise the  
pressure of the water.  
Pumping is accomplished by transforming the input energy - typically from an electric motor or  
from other sources such as high-pressure air.  
• The pump calculations in this section are for electrically driven rotodynamic pumps.  
To move water, a pump will need to overcome resistance due its density, gravitational force and  
friction.  
• This resistance is dependent on:  
Height the water needs to be raised. This height of the fluid in a container is referred to as  
head.  
Quantity of water involved  
21.1 Glossary of Pump Calculations Terms  
Force: In the English system force and weight are often used in the same way. The weight of the cubic  
foot of water is 62.4 pounds. The force exerted on the bottom of the one foot cube is 62.4 pounds. If we  
have two cubes stacked on top of one another, the force on the bottom will be 124.8 pounds.  
Pressure: Pressure is a force per unit of area, pounds per square inch or pounds per square foot are com-  
mon expressions of pressure.  
Head: Pressure is directly related to the height of a column of fluid. This height is called head or feet of  
head. Pressure and feet of head head are directly related - for every one foot of head there is a pressure of  
0.433 psi.  
0.433 psi  
ft (water column)  
1 ft (water column)  
2.31 psi  
Thus,  
or conversely  
Note: This pressure/head will include the height the water pumped and also the head associated with friction  
losses - energy loss because of the water moving through the pipe and fittings.  
 
 
326  
Chapter 21. Pumping Efficiencies and Costs  
Force  
Area  
Pressure =  
12"  
Pressure exerted by  
a 1ft column of water  
62.4 lb  
12in x 12in  
= 0.43 psi  
=
12"  
As 1ft3 of water weighs 62.4 lbs  
12"  
The pressure at the bottom of a container is affected only by the height of water in the container and not  
by the shape or the volume of the container. In the drawing below there are four containers all of different  
shapes and sizes. The pressure at the bottom of each is the same.  
The pressure exerted at the bottom of a tank is relative only to the head on the tank and not the volume  
of water in the tank. For example, below are two tanks each containing 5000 gallons. The pressure at the  
bottom of each is 22 psi. If half of the water were drained from the tanks the pressure at the bottom of the  
elevated tank would be 17.3 psi while the pressure at the bottom of the standpipe would be 11 psi.  
Velocity: Velocity is the speed that the water is moving along a pipe or through a basin. Velocity is  
usually expressed in feet per second, ft/sec.  
Flow: Flow is commonly expressed in gallons per minute (gpm) and/or cubic feet per second (cfs). There  
is a relationship between gallons per minute and cubic feet per second. One cubic foot per second is equal  
to 448.8 gallons per minute.  
1cfs = 448.8gpm  
21.1 Glossary of Pump Calculations Terms  
327  
Flow Equation; The basic equation for determining flow is as follows:  
Q = V ×A  
Where:  
Q = cfs ft3/sec  
V = ft/sec  
A = ft2  
Static Pressure: Static implies a non-moving condition. The pressure measured when there is no water  
moving in a line or the pump is not running is called static 32 pressure. This is the pressure represented by  
the gauges on the tanks in the discussion above.  
Dynamic Pressure: When water is allowed to run through a pipe and the pressure (called pressure head)  
measured at various points along the way we find that the pressure decreases the further we are from the  
sources.  
Headloss: The reason for this reduction in pressure is a phenomenon called headloss. Headloss is the  
loss of energy (pressure) due to friction. The energy is lost as heat.  
If the headloss in a certain pipe is 25 feet, it means the amount of energy required to overcome the fric-  
tion in the pipe is equivalent to the amount of energy that would be required to lift this amount of water  
straight in the air 25 feet.  
In a pipe, the factors that contribute to headloss include the following:  
• Roughness of pipe - If the roughness of a pipe were doubled the headloss would double.  
• Length of pipe - If the length of the pipe were doubled the headloss would double.  
• Diameter of pipe - If the diameter of a pipe were doubled the headloss would be cut in half  
Velocity of water - If the velocity of the water in a pipe were doubled the headloss would be in-  
creased by about four times. It should be apparent that velocity, more than any other single factor,  
affects headloss. To double the velocity we would have to double the flow in the line.  
Pumping System Components and Fittings - Each type of fitting has a specific headloss depending  
upon the velocity of water through the fitting. For instance the headloss though a check valve is  
two and one quarter times greater than through a ninety degree elbow and ten times greater than the  
headloss through an open gate valve.  
Static Head: Static head is the distance between the suction and discharge water levels when the pump is  
shut off.  
Suction Lift: Suction lift is the distance between the suction water level and the center of the pump  
impeller. This term is only used when the pump is in a suction lift condition. A pump is said to be in a  
suction lift condition any time the eye (center) of the impeller is above the water being pumped.  
Velocity Head: The amount of energy required to bring a fluid from standstill to its velocity. For a given  
quantity of flow, the velocity head will vary indirectly with the pipe diameter.  
Total Dynamic Head (TDH): The total energy needed to move water from the center line of a pump (eye  
of the first impeller of a lineshaft turbine) to some given elevation or to develop some given pressure. This  
includes the static head, velocity head and the headloss due to friction.  
328  
Chapter 21. Pumping Efficiencies and Costs  
Horsepower: Horsepower is a measurement of the amount of energy required to do work. Motors are  
rated in horsepower. The horsepower of an electric motor is called brake horsepower. The horsepower  
requirements of a pump are dependent on the flow and the total dynamic head. 33,000 foot pounds per  
minute of work is 1 horsepower.  
Suction Head: Suction head is the distance between the suction water level and the center of the pump  
impeller when the pump is in a suction head condition. A pump is said to be in a suction head condition  
any time the eye (center) of the impeller is below the water level being pumped.  
Velocity Head: Velocity head is the amount of energy required by the pump and motor to overcome  
inertia and bring the water up to speed. Velocity head is often shown mathematically as V2/2 g. (  
acceleration due to gravity 32.2ft/sec2 ).  
g is the  
Total Dynamic Head: Total dynamic head (TDH) is a theoretical distance. It is the static head, velocity  
head and headloss required to get the water from one point to another.  
The horsepower output of an electric motor is directly reflected to the amperage that the motor draws. Any  
increase in horsepower requirements will give a corresponding increase in amperage.  
Cavitation: Cavitation in pumps is the rapid creation and subsequent collapse of air bubbles occuring as  
a result of the inlet pressure falling below the design inlet pressure or when the pump is operating at a flow  
rate higher than the design flow rate. This collapse of the air bubbles typically manifests as a pinging or  
crackling noise. Cavitation is undesirable because it can damage the impeller, cause noise and vibration,  
and decrease pump efficiency.  
21.2 Pumping Rate Calculations  
For calculating volume pumped given the pump flow rate: Multiply the pump flow rate by the time  
interval  
Make sure:  
The time units - in the given time interval and in the pump flow rate match  
• For calculating time to pump a certain volume:  
Step 1. Calculate the total volume pumped  
Step 2. Divide the total volume by the pump flow rate  
Make sure:  
The volume units - in the volume that needs to be pumped and in the pump flow rate match  
The time unit in the pump flow rate needs to be converted to the time unit that you need the  
answer in  
 
21.3 Power Requirements for Pumping  
329  
21.3 Power Requirements for Pumping  
Where:  
Input Hp is the input power to the motor which produces the Output Hp or Brake Hp - the  
mechanical power which runs the pump.  
• The ratio of Output Hp and Input Hp is the motor efficiency - ηm.  
• The Output Hp is the input power (Brake Hp) to the pump to pump the water.  
Water Hp is the rate of energy transferred to the water being pumped and can be calculated by the  
formula:  
H Head of water (ft) Q Flow (GPM)  
3,960 (Conversion factor for converting GPMft to Hp)  
• The ratio of Output Hp and Water Hp is the pump efficiency - ηp.  
 
330  
Chapter 21. Pumping Efficiencies and Costs  
21.4 Summary of conversions and formulas  
Total dynamic head = Friction head + Static head  
1Hp = 0.746kW  
33,000 ft lb  
1Hp =  
min  
1Hp = 3,960 GPM ft  
1psi = 2.31 ft  
Water Hp = Flow Head  
Brake Hp = Input HpMotor e f ficiency  
Water Hp = Brake Hp Pump e f ficiency, and  
=Water Hp = Input Hp Motor e f ficiency Pump e f ficiency  
 
PUMPS & PUMPING  
1
2
3
Pump is a device for raising  
To move water, need to overcome  
This resistance is dependent on:  
or moving water or any other fluid.  
resistance due its density,  
• Height the water needs to be raised  
• Quantity of water involved  
gravitational force & friction.  
4
5
6
Power needs to be delivered to the pump so it can provide energy. Power = Energy per time  
Force needs to be exerted  
by the pump to overcome  
the resistance  
Pump will need to  
provided energy to raise  
the water  
Power Units  
ft-lb  
min  
KW  
Hp  
Understanding the concept of power:  
1Hp = 0.746 kW  
Energy = resistance force  
Energy = Force x Distance  
Force is the head which is measured in terms  
of the height of water - inches or feet  
A 150lb person climbing 50ft will expend  
7500 ft-lb of work (energy)  
Force = Mass x Acceleration  
Watt determined that one horse on an  
average could lift 330lbs 100ft in one minute  
(lbf)  
(lbm)  
Energy units  
ft-lb  
KWh  
Calories  
Hp-h  
1) Power requirement for climbing this in 5 minutes  
lbf = lbm  
(per definition)  
33,000 ft-lb  
1Hp =  
min  
ft-lb  
min  
7500  
5
ft-lb  
min  
= 1500  
1'=12"  
1'=12"  
8.34lb  
As 1 GPM (Water) =  
min  
= 0.045 Hp  
force  
area  
Pressure =  
GPM  
8.34  
lb  
min  
=
1'=12"  
2) Power requirement for climbing this in 1 minute  
33000 ft-lb  
min  
33000 ft x GPM  
8.34 min  
1Hp =  
=
7500  
ft-lb  
min  
= 0.23 Hp  
Pressure exerted by  
a 1ft column of water 12 in x 12 in  
62.4 lb  
=
= 0.43 psi  
1Hp = 3,960 GPM-ft  
so 0.43 psi = 1ft (water column)  
or 1 psi = 2.3ft  
1Hp is needed to raise one gallon of water 3,960  
ft in one minute  
Flow * Head  
Friction Head  
Head associated with energy  
losses due energy losses due  
to friction  
Input Hp  
$
Output Hp  
Brake Hp  
Water Hp  
m
Head  
Total Dynamic Head  
Total equivalent height  
of water column  
Static Head  
needed to pump  
Head associated with the height  
the water has to be raised  
Given  
Calculate  
What is the required water horsepower to pump 300 GPM if the suction head and discharge heads are  
constant at 20 ft and 120 ft respectively and the system head losses are 5 ft.  
Calculating Static Head  
Solution:  
300 GPM x (120-20+5 ft)  
=
7.95Hp  
GPM-ft  
3960  
Final Water Surface  
El = y  
Hp  
Given  
Calculate  
Find the input Hp Given the water Hp equals 8Hp and the pump and motor efficiencies are 80%  
and 50% respectively.  
Solution:  
Step 1: Find  
knowing  
&
Step 2: Find  
knowing  
=
&
Brake Hp  
16  
8Hp  
Brake Hp  
= Brake Hp = 16Hp  
0.8 =  
0.5 =  
Input Hp  
Input Hp  
16  
0.8  
Input Hp =  
20Hp  
=
El = x  
El = 0  
El = -x  
Suction  
lift (x)  
Given  
and existing  
& replacement  
Calculate  
savings  
Suction  
Head (x)  
An older motor which is 82% efficient is to be replaced by a new, 94% efficient motor. Calculate the  
annual savings for the new motor given the output horsepower from both motors is 180Hp and the elec-  
tricity cost is $0.075/kWh & the motor operates 24 hours per day throughout the year.  
Case A:  
Initial water level  
above pump  
Case B:  
Initial water level below  
pump  
180  
0.94  
180  
0.82  
Hp input to  
new motor  
Hp input to  
old motor  
=
= 191.5Hp  
=
= 218.5Hp  
$0.075  
Kwh  
$23,779  
Year  
365 x 24 hrs  
Annual  
Savings  
0.746 Kw  
=
= y-x  
= y-(-x)  
= y+x  
X
Static Head  
Static Head  
(218.5-191.5)Hp  
X
X
Year  
Hp  
“Demand Charge” ($/kw) is imposed on consumer by the utility company to compensate it for the design & upkeep of equipment to meet the peak power draw by the consumer. The demand charge is in addition to the regular energy  
consumption charge ($/kw). The demand charge is based upon highest power consumption over any 15 minute period during the billing cycle. Oversized and inefficient equipment (motors & pumps) would mean higher demand charges.  
21.5 Practice Problems  
333  
21.5 Practice Problems  
1. A reservoir is 40 feet tall. Find the pressure at the bottom of the reservoir.  
40ft×0.433psi/ft = 17.3psi  
2. Find the height of water in a tank if the pressure at the bottom of the tank is 12 psi.  
12psi÷0.433psi/ft = 27.7ft  
3. If a pump discharge pressure gauge read 10 psi, the height of the water corresponding to this pres-  
sure would be:  
2.31 ft  
10 psi×  
= 23.1 ft  
psi  
4. A pump is set to pump 5 minutes each hour. It pumps at the rate of 35 gpm. How many gallons of  
water are pumped each day?  
Solution:  
¨
35 gal sludge 5 min 24 hr  
4,200 gallons  
¨
=
¨
min  
hr  
day  
day  
¨
Example 2: A pump operates 5 minutes each 15 minute interval. If the pump capacity is 60 gpm,  
how many gallons are pumped daily?  
H¨  
¨
60 gal sludge 5 min  
min  
28,800 gal sludge  
¨H  
¨
1440  
¨
=
H¨  
min  
¨H  
15 min  
day  
day  
¨
5. Given the tank is 10ft wide, 12 ft long and 18 ft deep tank including 2 ft of freeboard when filled to  
capacity. How much time (minutes) will be required to pump down this tank to a depth of 2 ft when  
the tank is at maximum capacity using a 600 GPM pump  
Solution:  
2’ Freeboard  
16’ Water Depth (Initial)  
2’ Water Depth (Final)  
10’ Wide  
12’ Long  
Volume to be pumped=12 ft 10 ft (162) ft = 1,680ft3  
ft3  
gal  
3
1,680ft 7.48  
=⇒  
= 21min  
gal  
600  
min  
6. 1 MGD is pumped against a 14’ head. What is the water Hp? The pump mechanical efficiency is  
85%. What is the brake horsepower?  
water Hp = flow * head  
1,000,000 gal  
day  
1440 min  
Hp  
14 ft ∗  
= Water Hp = 2.46 Hp  
day  
3,960 GPM ft  
 
334  
Chapter 21. Pumping Efficiencies and Costs  
pump Hp = brake Hp * pump efficiency  
2.46  
0.85  
Brake Hp =  
= Brake Hp = 2.89Hp  
7. A 8 ft diameter cylindrical wetwell receives an average incoming flow if 135 gpm and is pumped  
down with a pump that delivers 450 gpm again a total dynamic head of 120 ft. The pump is con-  
trolled using two floats; a stop float located at 2.5 ft and a start float located at 16 ft. If the pump  
motor is rated at 88% and the pump at 77%, what is the monthly (30 days/month) for running this  
pump if power costs are $0.11/Kwh?  
When the pump is on, the volume of wetwell that will be pumped down with the 450 gpm pump  
and a 135 gpm flow to the wetwell:  
450 gal 135 gal  
315 gal  
min  
=
min  
min  
Minutes required to pump down the wetwell :  
7.48 gal  
min  
315 gal  
0.78582 (162.5) ft3 ∗  
= 16.1 min  
ft3  
Time to fill wetwell with pump off @135gal/min influent flow:  
7.48 gal  
min  
135 gal  
[0.78582 (162.5)] ft3 ∗  
= 37.6min  
ft3  
# of cycles per day:  
cycle  
(16.1+37.6) min  
1440 min 26.8 cycles  
=
day  
day  
# of hrs pump operational:  
16.1 min 26.8 cycles  
hrs  
60 min  
7.19 hours  
=
cycle  
day  
day  
Monthly electrical cost:  
450 gpm120 ft  
Hp  
0.746 kW 7.19hrs 30 days $0.11  
$356  
month  
=
0.880.77  
3,960 gpmft  
Hp  
day  
month  
kWh  
8. A 6-year old pump motor is to be replaced at a net cost of $15,800. The new motor, just like the old  
one, would run 65% of the time. Both existing and replacement motors would operate at 125 output  
Hp. The existing motor efficiency is 86% while the replacement motor would be guaranteed at 94%  
efficiency. Electricity currently averages $0.088 per kWh.  
(a) Calculate the energy cost savings per year (to the nearest dollar) if the existing motor is re-  
placed with the new motor (neglect any consideration of impact upon demand charges or interest  
on capital).  
(b) What is payback period to the nearest tenth of a year.  
Solution:  
21.5 Practice Problems  
335  
Calculate energy cost savings per year:  
125  
Input Hp for old motor:  
= 145.35Hp  
0.86  
125  
0.94  
Input Hp for old motor:  
= 132.98Hp  
Energy cost savings:  
0.746 kW (365240.65)hrs $0.088  
=
$4,624  
(145.35132.98)Hp∗  
Hp  
yr  
kWh  
yr  
Calculate payback:  
yr  
$15,800∗  
= 3.4yr  
$4,623.94  
9. A flow of 200 gpm is pumped against a total head of 4.0 feet. The pump is 78% efficient and the  
motor’ is 90% efficient. Calculate the input Hp.  
water Hp = flow * head  
Hp  
200GPM 4 ft ∗  
= 0.2Hp  
3,960GPM ft  
water Hp=brake Hp*pump efficiency, and  
brake Hp=input Hp*motor efficiency  
Therefore, water Hp=input Hp*motor efficiency*pump efficiency  
water Hp  
0.2  
input Hp=  
=
= 0.28Hp  
motor e f ficiencypump e f ficiency 0.90.78  
Chapter Assessment  
1. A 240 volt motor runs an average of 8 hours a day. If the electric meter registered 6,450 kilowatt  
hours for a 31-day month, what is the motor horsepower? [35 Hp]  
2. A pump is equipped with a pressure gauge in the discharge pipe that reads 100 psi. The total dis-  
charge head in feet would be?  
2.31ft water  
100psi∗  
= 231 ft water  
psi  
3. 900 GPM pump is pumped against a 12 ft head. What is the water Hp water Hp = flow * head  
Hp  
900GPM 12 ft 3,960GPMft = 2.7Hp  
4. A 50 ft3/sec flow is pumped against a head of 8 feet. What is the water Hp  
water Hp = flow * head  
5
0
7.48gal  
ft  
Hp  
60sec  
min  
ft3sec8 ft ∗  
= 45.4Hp  
3
3,960GPMft  
5. 1 MGD is pumped against a 14’ head. What is the water Hp? The pump mechanical efficiency is  
85%. What is the brake horsepower?  
water Hp = flow * head  
1,000,000gal  
day  
Hp  
1440min 14 ft 3,960GPMft = Water Hp = 2.46Hp  
day  
pump Hp = brake Hp * pump efficiency  
2.46  
0.85  
brake Hp =  
= Brake Hp = 2.89Hp  
6. A flow of 2.5 MGD is being lifted 10 feet and then pumped up another 120 feet to a storage reser-  
voir. Calculate the pump output power required to lift this water. Ignore friction losses.  
water Hp = flow * head  
2,500,000gal  
day  
Hp  
1440min (120+10) ft 3,960GPMft = Water Hp = 57Hp  
day  
338  
Chapter 21. Pumping Efficiencies and Costs  
7. A pump is pumping 400 gpm. The suction pressure gauge indicates a pressure of 5 ft and the pump  
discharge pressure gauge indicates a pressure of 100 ft. If the pump brake horse power is 12 hp,  
what is the pump efficiency  
water Hp = flow * head  
Hp  
400gpm(1005) ft 3,960gpmft = 9.6Hp  
pump efficiency - ηp=9.6Hp 100 = 79%  
12Hp  
8. A flow of 200 gpm is pumped against a total head of 4.0 feet. · The pump is 78% efficient and the  
motor’ is 90% efficient. Calculate the input Hp.  
water Hp = flow * head  
Hp  
200GPM 4 ft 3,960GPMft = 0.2Hp  
water Hp=brake Hp*pump efficiency, and  
brake Hp=input Hp*motor efficiency  
Therefore, water Hp=input Hp*motor efficiency*pump efficiency  
water Hp  
e f ficiencypump e f ficiency  
0.2  
0.90.78  
input Hp=motor  
=
= 0.28Hp  
21.5 Practice Problems  
339  
9. 500,000 gpd of secondary· effluent is pumped to a storage pond for reuse as golf course irrigation  
water. The water is lifted 12 feet in the plant, and then pumped up another 75 feet to the storage  
pond. Friction losses are assumed to be 10% of the static head. Assuming the pump efficiency of  
70% and a motor efficiency of 92% and an electrical cost of $0.0725 per KWh, calculate the daily  
cost of pumping this water.  
Solution:  
water Hp = flow * head  
500,000gal  
day  
Hp  
1440min (87 ft static head +870.1 ft friction head)3,960GPMft  
day  
= 8.39water Hp  
water Hp  
e f ficiencypump e f ficiency  
8.39  
0.920.70  
input Hp=motor  
=
= 13Hp  
$16.87  
day  
0.746kW 24hrs $0.0725  
Electrical cost=13Hp∗  
=
kWh  
Hp  
day  
10. A pump motor (93% efficient) generates an output of 130 HP and runs 75% of the time. Electricity  
costs an average of 8.455 cents per kilowatt-hour. What is the monthly cost of operating this pump  
in $ per month?  
$4,761  
month  
130Hp  
0.93  
30days  
month  
0.746kW 24hrs  
$0.08455  
kWh  
0.75∗  
=
Hp  
day  
340  
Chapter 21. Pumping Efficiencies and Costs  
11. A wet well is 8 ft x 8 ft x 16.5 ft deep and receives a continuous flow of 310,000 gpd. A 500 gpm  
pump draws down 12 feet of water each pumping cycle. The motor that drives the pump draws 52.5  
Hp when it pumps. The cost of electricity is $0.0755 per kilowatt - hour. Calculate  
(a) the time it takes to pump down the wet well, and  
(b) The daily electrical energy cost for this pump.  
Solution:  
310,000gal  
day  
215.3gal  
min  
@
=
1440min  
day  
Volume of wetwell that will be pumped down with the 500 gpm pump and a 215.3 gpm flow to the  
wetwell:  
500gal  
min  
215.3gal  
min  
284.7gal  
min  
=
Minutes required to pump down the wetwell :  
7.48gal  
ft  
min  
8812 ft3 ∗  
= 20.2min  
284.7gal  
3
Time to fill wetwell with pump off @215.3gal/min influent flow:  
7.48gal  
ft  
min  
8812 ft3 ∗  
= 26.7min  
215.3gal  
3
# of cycles per day:  
cycle  
30.7cycles  
1440min  
=
day  
day  
(20.2+26.7)min  
# of hrs pump operational:  
30.7cycles  
20.2min  
hrs  
10.33hours  
=
60min  
cycle  
day  
day  
Daily electrical cost:  
$30.54  
day  
0.746kW 10.33hrs $0.0755  
52.5Hp∗  
=
kWh  
Hp  
day  
21.5 Practice Problems  
341  
12. A 6-year old pump motor is to be replaced at a net cost of $15,800. The new motor, just like the old  
one, would run 65% of the time. Both existing and replacement motors would operate at 125 output  
Hp. The existing motor efficiency is 86% while the replacement motor would be guaranteed at 94%  
efficiency. Electricity currently averages $0.088 per kWh.  
(a) Calculate the energy cost savings per year (to the nearest dollar) if the existing motor is re-  
placed with the new motor (neglect any consideration of impact upon demand charges or interest  
on capital).  
(b) What is payback period to the nearest tenth of a year. Energy cost savings per year:  
125  
Input Hp for old motor:  
Input Hp for old motor:  
Energy cost savings:  
= 145.35Hp  
= 132.98Hp  
0.86  
125  
0.94  
$4,623.94  
(365240.65)hrs  
0.746kW  
Hp  
$0.088  
(145.35132.98)Hp∗  
=
kWh  
yr  
yr  
Calculate payback:  
yr  
$15,800$4,623.94 = 3.4yr  
13. A 8 ft diameter cylindrical wetwell receives an average incoming flow if 135 gpm and is pumped  
down with a pump that delivers 450 gpm again a total dynamic head of 120 ft. The pump is con-  
trolled using two floats; a stop float located at 2.5 ft and a start float located at 16 ft. If the pump  
motor is rated at 88% and the pump at 77%, what is the monthly (30 days/month) for running this  
pump if power costs are $0.11/Kwh? (Ans: $356/month)  
Volume of wetwell that will be pumped down with the 450 gpm pump and a 135 gpm flow to the  
wetwell:  
450gal  
min  
135gal  
min  
315gal  
=
min  
Minutes required to pump down the wetwell :  
7.48gal  
ft  
min  
0.78582 (162.5) ft3 ∗  
= 16.1min  
315gal  
3
Time to fill wetwell with pump off @135gal/min influent flow:  
7.48gal  
ft  
min  
135gal  
[0.78582 (162.5)] ft3 ∗  
= 37.6min  
3
# of cycles per day:  
cycle  
26.8cycles  
1440min  
=
day  
day  
(16.1+37.6)min  
# of hrs pump operational:  
26.8cycles  
16.1min  
hrs  
7.19hours  
day  
=
cycle  
day  
60min  
Daily electrical cost:  
$356  
day  
450gpm120 ft  
0.880.77  
Hp  
30days  
month  
0.746kW 7.19hrs  
$0.11  
=
kWh  
3,960gpmft  
Hp  
day  
14. A pump operating at 80% effieciency generates an water Hp of 60 HP and runs 75% of the time.  
Assuming the pump motor is 90% efficent and electricity costs an average of $0.0821 per kilowatt-  
hour. The monthly (30 days) cost of operating this pump is:  
342  
Chapter 21. Pumping Efficiencies and Costs  
$2,756  
month  
60Hp  
0.900.80  
30days  
month  
0.746kW 24hrs  
$0.0821  
kWh  
0.75∗  
=
Correct Answer  
Hp  
day  
$2,480  
month  
60Hp  
0.8  
30days  
0.746kW 24hrs  
$0.0821  
0.75∗  
0.75∗  
=
=
Hp  
day  
month  
kWh  
$2,205  
month  
60Hp  
0.90  
30days  
month  
0.746kW 24hrs  
$0.0821  
kWh  
Hp  
day  
$3,675  
month  
60Hp  
0.900.80  
30days  
month  
0.746kW 24hrs  
$0.0821  
kWh  
=
Hp  
day  
15. A 8 ft diameter cylindrical wetwell receives an average incoming flow if 135 gpm and is pumped  
down with a pump that delivers 450 gpm again a total dynamic head of 120 ft. The pump is con-  
trolled using two floats; a stop float located at 2.5 ft and a start float located at 16 ft. If the pump  
motor is rated at 88% and the pump at 77%, what is the monthly (30 days/month) for running this  
pump if power costs are $0.11/Kwh? (Ans: $356/month)  
16. A 8 ft diameter cylindrical wetwell receives an average incoming flow if 135 gpm and is pumped  
down with a pump that delivers 450 gpm again a total dynamic head of 120 ft. The pump is con-  
trolled using two floats; a stop float located at 2.5 ft and a start float located at 16 ft. If the pump  
motor is rated at 88% and the pump at 77%, what is the monthly (30 days/month) for running this  
pump if power costs are $0.11/Kwh? (Ans: $356/month)  
17. If a wasting pump has a fixed pump rate of 250 GPM, and your calculation indicates you must  
waste 126,000 gallons, what hourly cycle rate do you set the timer?  
a. Turn pump on 21 minutes every day  
b. Turn pump on 504 minutes every hour  
c. Turn pump on 42 minutes every day  
d. Turn pump on 21 minutes every hour  
Solution:  
21 min  
126,000 gal  
day  
min  
hr  
min  
=
24 hrs 250 gal  
=
day  
hr  
18. A pump operating at 80% efficiency generates an water Hp of 60 HP and runs 75% of the time.  
Assuming the pump motor is 90% efficent and electricity costs an average of $0.0821 per kWh.  
The monthly (30 days) cost of operating this pump is:  
*a. $2,756  
b. $2,480  
c. $2,205  
d. $3,675  
19. How long will it take to pump down 25 feet of water in a 110 ft diameter cylindrical tank when  
using a 1420 gpm pump.  
a. 26 hours and 56 minutes  
*b. 26 hours and 33 minutes  
c. 2 hours and 47 minutes  
d. 12 hours and 36 minutes  
20. How long will it take to pump down 25 feet of water in a 110 ft diameter cylindrical tank when  
using a 1420 gpm pump.  
a. 26 hours and 56 minutes  
*b. 26 hours and 33 minutes  
c. 2 hours and 47 minutes  
21.5 Practice Problems  
343  
d. 12 hours and 36 minutes  
21. A positive displacement pump should be started up with the discharge valve closed in order to  
avoid any problems with "air lock".  
a. True  
*b. False  
22. Brake horse power is the input power to the motor  
a. True  
*b. False  
23. Cost of electrical usage for pumps is based upon kilowatt per hour  
a. True  
*b. False  
24. A centrifugal pump can be used to pump sludge.  
*a. True  
b. False  
25. Variable speed sludge pumps may be used to keep the density of the sludge nearly constant.  
*a. True  
b. False  
The keeping of records of plant operation and maintenance, even when a portion of the plant is  
temporarily out of balance, is an integral part of good operation.  
*a. True  
b. False  
26. A positive displacement pump could be damaged if it is started with the discharge valve closed.  
*a. True  
b. False  
27. The common unit of power or rate of doing work is horsepower. This· is equal to 746 ft. lbs/min.  
*a. True  
b. False  
28. Total dynamic head is the sum of the suction head and the discharge head minus the friction head  
a. True  
*b. False  
29. Water power is the output power of the pump  
*a. True  
b. False  
30. A positive displacement pump should be started up with the discharge valve closed in order to  
avoid any problems with "air lock".  
a. True  
*b. False  
31. Brake horse power is the input power to the motor  
a. True  
*b. False  
32. Cost of electrical usage for pumps is based upon kilowatt per hour  
a. True  
*b. False  
33. The common unit of power or rate of doing work is horsepower. This· is equal to 746 ft. lbs/min.  
*a. True  
344  
Chapter 21. Pumping Efficiencies and Costs  
b. False  
34. Total dynamic head is the sum of the suction head and the discharge head minus the friction head  
a. True  
*b. False  
35. Water power is the output power of the pump  
*a. True  
b. False  
36. What is the vertical distance between the elevation of the free water surface at the suction and that  
of the free water surface at the discharge of a pump called?  
a. Discharge head.  
b. Dynamic head.  
c. Velocity head.  
*d. Static head.  
37. If a wasting pump has a fixed pump rate of 250 GPM, and your calculation indicates you must  
waste 126,000 gallons, what hourly cycle rate do you set the timer?  
a. Turn pump on 21 minutes every day  
b. Turn pump on 504 minutes every hour  
c. Turn pump on 42 minutes every day  
*d. Turn pump on 21 minutes every hour  
38. A positive displacement pump is connected to a 25’ wide x 125’ long x 12’ side water depth aer-  
obic digester. How long will it take to empty the contents of the digester if the pump rate is 225  
gallons per minute?  
a. 15.3 hours  
b. 2.8 hours  
*c. 20.8 hours  
d. 15.6 hours  
39. A centrifuge is fed sludge with a concentration of 3.4% solids. If the sludge feed rate is set at 50  
gallons per minute, what is the centrifuge loading rate in pounds per hour?  
a. 763 lbs/hour  
*b. 850 lbs/hour  
c. 735 lbs/hour  
d. 960 lbs/hour  
40. Calculate the surface loading rate for a treatment plant with 4 clarifiers each with a 100 foot diame-  
ter. The plant has an influent flow of 35 MGD.  
a. 279 gal/sq ft  
b. 950 gal/sq ft  
c. 4,459 gal/sq ft  
*d. 1,115 gal/sq ft  
41. Calculate the flow velocity in feet/minute if 7.5 MGD of flow passes through a channel that is 3’  
wide x 4’ deep, and the depth of flow is 15 inches.  
*a. 186 ft/min  
b. 58 ft/min  
c. 202 ft/min  
d. 46.5 ft/min  
42. Determine the pounds per day of primary solids removed at a plant with a flow rate of 1.5 MGD  
21.5 Practice Problems  
345  
and the following data:  
Clear Waters Fall 2013  
Influent TSS = 250 mg/L  
Primary Effluent TSS = 150 mg/L,  
Final Effluent TSS = 12 mg/L  
a. 1,101 lbs/day  
*b. 1,251 lbs/day  
c. 982 lbs/day  
d. 2,977 lbs/day  
43. A sewage pump is located above the wet well which is 8 feet deep and the pump is pumping to an  
above ground clarifier with 12 feet depth of water. The pump manufacturer has given you the pump  
characteristics curve which shows Total Dynamic Head vs. flow rates. If the operating wet well  
water depth is 6 feet, what is the total dynamic head in order to determine pumping rate from the  
chart? Assume the top of the wet well and the bottom of the clarifier are at the same elevation.  
a. 12 feet  
b. 20 feet  
*c. 14 feet  
d. 10 feet  
44. A sewage pump is located above the 8-foot diameter wet well which is 8 feet deep and the pump is  
pumping to an above ground clarifier. The flow meter on the pump is not operating and you want to  
calculate the pumping rate by measuring the drop in wet well water level during when inflow to wet  
well is minimal? If the drop in water level in one minute is 2 feet, what is the approximate pumping  
rate in gallons per minute?  
a. 250 GPM  
b. 375 GPM  
c. 500 GPM  
*d. 750 GPM  
22. Water Math  
22.1 Fractions  
A fraction is defined as part of whole. If in a class there are 20 male students and 30 male students,  
20  
the fraction of male students is or .  
50  
• It is composed of three items: two numbers and a line.  
2
5
The number on the top is called the numerator, the number on the bottom is called the denominator,  
and the line in between them means to divide.  
Numerator  
3
4
Divide →  
Denominator  
A proper fraction is a fraction that has no whole number part and its numerator is smaller than its  
denominator. An improper fraction is a fraction that has a larger numerator than denominator and it  
represents a number greater than one.  
1 5 11  
Proper Fraction Examples:  
, ,  
2 8 12  
12 5  
Improper Fraction Examples:  
,
2
2
Any whole number can be expressed as a fraction by placing a "1" in the denominator. For exam-  
ple:  
2
45  
1
2 is the same as and 45 is the same as  
1
Only fractions with the same denominator can be added/subtracted, and only the numerators are  
added/subtracted. For example:  
1
8
3
8
4
8
7
8
3
8
4
8
+
=
and,  
=
 
 
348  
Chapter 22. Water Math  
• A fraction combined with a whole number is called a mixed number. For example:  
1
8
2
3
3
4
1
2
17  
32  
4 , 16 , 8 , 45 and, 12  
These numbers are read, four and one eighth, sixteen and two thirds, eight and three fourths, forty-  
five and one half, and twelve and seventeen thirty seconds.  
Mized numbers  
A fraction can be changed by multiplying the numerator and denominator by the same number.  
This does not change the value of the fraction, only how it looks. For instance:  
1
2
1
2
2
2
2
4
is the same as  
×
which is  
17  
4
• Steps to convert  
to a mixed number:  
Step 1. How many times can 4 fit into 17? 4 because 4×4=16. Thus, 4 becomes the whole number  
part  
Step 2. How much is left over in the numerator? 1 because 1716 = 1. Thus, 1 becomes the numera-  
tor of the fractional part  
17  
4
1
4
Step 3.  
= 4  
To turn a mixed number into an improper fraction, multiply the whole number part by the denomi-  
nator and add the numerator. This becomes the new numerator over the original denominator.  
Example: Converting 1.5 feet to fraction  
1
1.5 ft = 1  
2
1
2
12+1 2+1  
3
2
1
=
=
=
2
2
A mixed value - say a circumference is given in feet and fraction of feet (say 7 3/4), needs to be  
converted to a fraction for calculation purposes.  
22.2 Ratio  
• Ratio is used for comparing the size of two or more quantities.  
5
10  
Say if there are 10 red cubes and 5 pink marbles in a bag, the ratio  
and red cubes. It can also be represented by 5:10.  
is the ratio of pink marbles  
• 5 lbs of chemical in 10 gallons solution is a ratio. So is 30 miles per gallon.  
• Unlike fractions, ratio does not compare things that have the same units.  
22.3 Proportion  
• Two quantities are said to be in proportion if one changes, the other changes in a specific way.  
Two quantities are said to be directly proportional, if the increase in one will increase the other  
value proportionally.  
x
y
Thus, if two quantities x and y are directly proportional, its ratio will be a fixed value. Thus  
x
y
x1  
y1  
for x1 and y1 different values of x and y respectively will be related by the equation  
=
.
 
 
22.3 Proportion  
349  
This relationship is useful for calculating unknown values in water treatment calculations as  
in the following example:  
Knowing 200 lbs of bleach is needed to disinfect 5 MG of water at a treatment plant, calcu-  
late the lbs of bleach required to disinfect 3.2 MGD flow.  
200 pounds bleach  
The ratio  
or 40 lbs bleach per MG is a constant. Using this known  
5MG  
proportion the lbs of bleach is needed to disinfect 3.2 MG at this plant can be calculated as  
follows by setting up the equation as:  
40 pounds bleach  
X
=
where X is the unknown lbs of bleach that is required to  
MG  
3.2 MG  
disinfect the 3.2 MG flow.  
3.240  
X can be calculated by cross multiplying the above equation: X =  
= 128 lbs bleach  
1
Two quantities are said to be inversely proportional if the increase in one will decrease the other  
value proportionally.  
Thus, if two quantities x and y are inversely proportional, its product xy will be a fixed value  
and different values of x and y respectively will be related by the equation xy = x1 y1.  
Examples of inversely proportional relationship include:  
Labor hours required to perform a certain task or time required to pump down a wetwell  
depending on the size of the pump. An increase in assignment of labor hours will reduce  
the time required to perform the task  
*
Using a larger pump will reduce the time to pump down the wetwell.  
In the Pounds formula:  
*
*
lbs  
day  
mg  
l
lbs or  
= Concentration  
8.34volume(MG) or Flow(MGD)  
for the same lbs or lbs/day, concentration varies inversely with volume or flow. Thus, for  
a certain pounds added, the concentration will go down if the flow increases and vice  
versa.  
In the flow equation, Q=V*A, for the same flow (Q), velocity (V) and surface area (A)  
are inversely related. If Q is remaining the same, an increase in surface area will reduce  
the velocity and vice versa.  
*
Additionally, for a flow through a pipe as the surface area of the pipe is proportional to  
the square of the diameter, the velocity in the pipe is inversely proportional to the square  
of the diameter.  
For a constant Q: V A = V1 A1 or V D2 = V1 D12  
Application of inversely proportional relationship in water related calculation can be demon-  
stration with the following example:  
If it takes 20 minutes to pump a wet well down with one pump pumping at 125 gpm, then  
how long will it take if a 200 gpm pump is used?  
As this is an inversely proportional relationship ( a larger pump will reduce the time re-  
quired):  
(20minutes125gpm) = (Xminutes200gpm)  
where X is the unknown time to pump down the wetwell with the 200 gpm pump.  
350  
Chapter 22. Water Math  
20125  
Solving for X: X =  
= 12.5 minutes  
200  
22.4 Decimals & Powers of Ten  
A decimal is composed of two sets of numbers: the numbers to the left of the decimal are whole  
numbers, and numbers to the right of the decimal are parts of whole numbers, a fraction of a num-  
ber.  
The term used to express the fraction component is dependent on the number of characters to the  
right of the decimal.  
The first character after the decimal point is tenths: 0.1 - tenths  
The second character is hundredths: 0.01 - hundredths  
The third character is thousandths: 0.001 - thousandths  
• Powers of 10 notation enables us to work with these very large and small quantities efficiently.  
In water math, the most common application of this concept is related to parts per million (ppm) or  
parts per billion (ppb).  
1 million - 1,000,000 can be represented as 106. Likewise, 1 billion - 1,000, 000,000 can be repre-  
sented as 109  
• The sequence of powers of ten can also be extended to negative powers.  
• 1 part per million (1/1,000,000) can be written as 106  
Name  
one  
Power  
100  
Number  
SI symbol SI prefix  
1
ten  
101  
102  
103  
106  
109  
101  
102  
10  
da (D)  
deca  
hecto  
kilo  
hundred  
thousand  
million  
billion  
tenth  
100  
h (H)  
1,000  
k (K)  
M
G
1,000,000  
1,000,000,000  
0.1  
mega  
giga  
d
deci  
hundredth  
thousandth 103  
0.01  
c
centi  
milli  
micro  
nano  
0.001  
m
µ
millionth  
billionth  
106  
109  
0.000 001  
0.000 000 001  
n
22.5 Rounding and Significant Digits  
Significant digits (also called Significant Figures) are digits which give us useful information about  
the accuracy of a measurement and are related to rounding.  
This concept is used to determine the direction to round a number (answer). The basic idea is that  
no answer can be more accurate than the least accurate piece of data used to calculate the answer.  
• Significant digits is the count of the numerals in a measured quantity (counting from the left) whose  
values are considered as known exactly, plus one more whose value could be one more or one less.  
• Rules for determining the number of significant digits:  
 
 
22.5 Rounding and Significant Digits  
351  
1. All nonzero digits are significant:  
1.234 g has 4 significant figures, and 1.2 g has 2 significant figures.  
2. Zeroes between nonzero digits are significant: 1002 kg has 4 significant figures, 3.07 mL has  
3 significant figures.  
3. Zeroes to the left of the first nonzero digits are not significant; such zeroes merely indicate  
the position of the decimal point: 0.001 C has only 1 significant figure, 0.012 g has 2 signifi-  
cant figures.  
4. Zeroes to the right of a decimal point in a number are significant: 0.023 mL has 2 significant  
figures, 0.200 g has 3 significant figures.  
5. When a number ends in zeroes that are not to the right of a decimal point, the zeroes are not  
necessarily significant: 190 miles may be 2 or 3 significant figures, 50,600 calories may be  
3, 4, or 5 significant figures. The potential ambiguity in the last rule can be avoided by the  
use of standard exponential, or ”scientific,” notation. For example, depending on whether 3,  
4, or 5 significant figures is correct, we could write 50,600 calories as: 5.06104 calories (3  
significant figures) 5.060 104 calories (4 significant figures), or 5.0600 104) calories (5  
significant figures).  
• Examples of significant figures:  
1000 has one significant digit: only the 1 is interesting (only it tells us anything specific); we don’t know anything for sure about the hundreds,  
tens, or units places; the zeroes may just be placeholders; they may have rounded something off to get this value.  
1000.0 has five significant digits: the ".0" tells us something interesting about the presumed accuracy of the measurement being made;  
namely, that the measurement is accurate to the tenths place, but that there happen to be zero tenths.  
0.00035 has two significant digits: only the 3 and 5 tell us something; the other zeroes are placeholders, only providing information about  
relative size.  
0.000350 has three significant digits: the last zero tells us that the measurement was made accurate to that last digit, which just happened to  
have a value of zero.  
1006 has four significant digits: the 1 and 6 are interesting, and we have to count the zeroes, because they’re between the two interesting  
numbers.  
560 has two significant digits: the last zero is just a placeholder.  
560. : notice that "point" after the zero! This has three significant digits, because the decimal point tells us that the measurement was made to  
the nearest unit, so the zero is not just a placeholder.  
560.0 has four significant digits: the zero in the tenths place means that the measurement was made accurate to the tenths place, and that there  
just happen to be zero tenths; the 5 and 6 give useful information, and the other zero is between significant digits, and must therefore also be  
counted.  
• Addition and Subtraction  
When you are adding or subtracting a bunch of numbers and need to be concerned with  
significant figures, first add (or subtract) the numbers given in their entire format, and then  
round the final answer. When rounding the final answer after adding or subtracting, the answer  
must be written with the same significant figures as the least accurate decimal place given.  
Example: 13.214 + 234.6 + 7.0350 + 6.38  
13.214 + 234.6 + 7.0350 + 6.38 = 261.2290  
*
*
234.6 is only accurate to the tenths place making it the least accurate number. My an-  
swer must be rounded to the same place as the least accurate number:  
261.2290 rounds to 261.2 (one decimal place)  
*
• Multiplication and Division  
When multiplying or dividing multiple numbers you would do these calculations as normal.  
352  
Chapter 22. Water Math  
When the answer must be written in the appropriate significant figure your answer must  
round to the same number of significant figures as the least number of significant figures.  
Example 1: Simplify, and round, to the appropriate number of significant digits  
16.235 x 0.217 x 5  
Step 1. First off, 5 has only one significant figure, thus the final answer needs to be rounded to  
one significant digit  
Step 2. 16.235 x 0.217 x 5 = 17.614975  
Step 3. To round 17.614975 to one digit. I’ll start with the 1 in the tens place. Immediately to  
its right is a 7, which is greater than 5, so 1 is rounded up to 2, and then replacing the 7  
with a zero, and dropping the decimal point and everything after it.  
Step 4. 17.614975 rounds to 20  
Example 2: Simplify, and round, to the appropriate number of significant figures  
0.00435 x 4.6  
Step 1. 4.6 has only 2 significant figures, so the final answer should be rounded to two signifi-  
cant figures.  
Step 2. 0.00435 x 4.6 = 0.02001  
Step 3. 0.02001 would round to 0.020, which has 2 significant figures (0.020). The answer  
cannot be 0.02, because that value would have only one significant figure.  
A number is rounded off by dropping one or more numbers from the right and adding zeroes, if  
necessary, to maintain the decimal point.  
If the last figure dropped is 5 or more, increase the last retained figure by 1. If the last digit dropped  
is less than 5, do not increase the last retained figure.  
22.6 Averages  
Also known as arithmetic mean, this value is arrived at by adding the quantities in a series and  
dividing the total by the number in the series.  
Example 1: Find the average of the following series of numbers: 12,8,6,21,4,5 , 9 , and 12.  
Adding the numbers together we get 77.  
There are 8 numbers in this set.  
Divide 77 by 8.  
77  
8
= 9.6 is the average of the set  
Example 2: Find the average of the set of daily turbidity data - 0.3,0.4,0.3,0.1,and 0.8  
The total is 1.9.  
There are 5 numbers in the set.  
Therefore:  
1.9  
= 0.38, rounding off = 0.4  
5
22.7 Working with Percent  
• Percent expresses portions of the whole.  
The whole is considered as 1 or 100% and a part of the whole can be expressed as a percent. Example:  
If a tank is 1/2 full, we say that it contains 50% of the original solution.  
• Percentage is written as a whole number with a % sign after it.  
 
 
22.7 Working with Percent  
353  
• In a calculation, percent is expressed as a decimal.  
• The decimal form of a percent value is obtained by dividing the percent by 100.  
Example: 11% is expressed as the decimal 0.11, since 11% is equal to 11/100. This decimal is  
obtained by dividing 11 by 100.  
• To determine what percentage a part is of the whole, divide the part by the whole.  
Example: There are 80 water meters to read, Jim has finished 24 of them. What percentage of the  
meters have been read?  
24÷80 = 0.30  
The 0.30 is converted to percent by multiplying the answer by 100.  
0.30×100 = 30%  
Thus 30% of the 80 meters have been read.  
To find the percentage of a number, multiply the number by the decimal equivalent of the percentage  
given in the problem.  
Example:  
What is 28% of 286?  
Step 1. Change the 28% to a decimal equivalent:  
28%÷100 = 0.28  
Step 2. Multiply 286×0.28 = 80  
Thus 28% of 286 is 80.  
To increase a value by a percent, add the decimal equivalent of the percent to " 1 " and multiply it  
times the number.  
Example: A filter bed will expand 25% during backwash. If the filter bed is 36 inches deep, how  
deep will it be during backwash?  
Step 1. Change the percent to a decimal.  
Step 2. Add the whole number 1 to this value.  
Step 3. Multiply times the value.  
25%÷100 = 0.25  
1+0.25 = 1.25  
36 in ×1.25 = 45 inches  
22.7.1 Percentage Concentrations  
Above concepts are used for chemicals such as fluoride and hypochlorites - the strength of the product as  
used is commonly expressed as a percentage.  
Example 1: A chlorine solution was made to have a 4% concentration. It is often desirable to determine  
this concentration in mg/L. This is relatively simple: the 4% is four percent of a million.  
 
354  
Chapter 22. Water Math  
To find the concentration in mg/L when it is expressed in percent, do the following:  
1. Change the percent to a decimal.  
4%÷100 = 0.04  
3. Multiply times a million.  
0.04×1,000,000 = 40,000mg/L  
We get the million because a liter of water weighs 1,000,000mg.1mg in 1 liter is 1 part in a million parts (  
ppm).1% = 10,000mg/L.  
Example 2: How much 65% calcium hypochlorite is required to obtain 7 pounds of pure chlorine?  
65% implies that in every lb of calcium hypochlorite has 65% lbs of available chlorine.  
0.65 lbs available chlorine  
lb of calcium hypochlorite  
lb of calcium hypochlorite  
0.65 lbs available chlorine  
Therefore,  
or conversely  
(
(
(
(
(
(
(
lb of calcium hypochlorite 7 lb of available chlorine  
(
(
(
=lbs calcium hypchlorite required =  
(
(
(
(
(
(
0.65 lbs available chlorine  
(
(
(
= 10.8 lbs of calcium hypochlorite with 65%available chlorine is required  
22.8 Area & Volume  
Example 1: The floor of a rectangular building is 20 feet long by 12 feet wide and the inside walls are 10  
feet high. Find the total surface area of the inside walls of this building  
Solution:  
 
22.8 Area & Volume  
355  
Ceiling=W*L  
Wall - L*H  
Wall - W*H  
Floor=W*L  
W*H  
Height=10’  
Wall - L*H  
Length=20’  
Width=12’  
2 Walls W*H + 2 Walls L*H= 21210 ft2 +22010 ft2  
= 240+400 = 640ft2  
2 Walls W*H + 2 Walls L*H + Floor + Ceiling= 21210 ft2 +22010 ft2 +21220 ft2  
= 240+400+480 = 1,120ft2  
Example 2: How many gallons of paint will be required to paint the inside walls of a 40 ft long x 65  
ft wide x 20 ft high tank if the paint coverage is 150 sq. ft per gallon. Note: We are painting walls only.  
Disregard the floor and roof areas.  
Solution:  
Wall - L*H  
W*H  
Height=20’  
Wall - W*H  
Wall - L*H  
Length=40’  
Width=65’  
2 Walls W*H + 2 Walls L*H = 26520 ft2 +24020 ft2 = 2,600+1,600 = 4,200ft2  
ft2  
gal  
4,200ft2  
=@150  
paint coverage →  
= 28 gallons Example 3: What is the circumference of  
ft2  
150  
gal  
a 100 ft diameter circular sedimentation tank?  
Solution:  
Circumference = π D = 3.14100ft = 314ft  
Example 4: If the surface area of a clarifier is 5,025ft2, what is its diameter?  
Solution:  
π
Sur face area = D2 =5025( ft2) = 0.785D2( ft2)  
4
5025  
0.785  
=D2 =  
=D = 6401.3 = 80ft  
Example 5: How many gallons of water would 600 feet of 6-inch diameter pipe hold, approximately?  
Solution:  
Diameter=6"  
π
4
Length=600’  
Volume = D2 ∗  
2
6
12  
gallons  
L = 0.785∗  
600 ft3 7.48  
= 881 gallons  
ft3  
356  
Chapter 22. Water Math  
22.9 Flow and Velocity  
• Flow Rate - Q (volume/time) = velocity (distance or length traveled /time) * surface area  
Velocity is the speed at which the water is flowing. It is measured in units of length/time – ft./sec.  
Velocity of water flowing through can be calculated by dividing the flow rate by area of the flow  
stream.  
volume or cubic length  
time  
sur face area in the direction of flowsquare length  
flow rate(  
)
length  
time  
Velocity  
=
For a flow in a channel:  
For a flow in a pipe:  
Example 1: If a chemical is added in a pipe where water is flowing at a velocity of 3.1 feet per second,  
how many minutes would it take for the chemical to reach a point 7 miles away?  
Note - we want the answer in minutes  
1 sec 5280 ft  
min  
60sec  
Min =  
7miles∗  
= 199min  
3.1 ft  
mile  
Example 2: Find the flow in cfs in a 6 -inch line, if the velocity is 2 feet per second.  
1. Determine the cross-sectional area of the line in square feet. Start by converting the diameter of the  
pipe to inches.  
The diameter is 6 inches: therefore, the radius is 3 inches. 3 inches is 3/12 of a foot or 0.25 feet.  
2. Now find the area in square feet.  
A = π ×r2  
A = π × 0.25ft2  
A = π ×0.0625ft2  
A = 0.196ft2  
 
22.10 Concentration  
357  
Or  
A = 0.785×D2  
A = 0.785×0.52  
A = 0.785×.05×.05  
A = 0.196ft2  
3. Now find the flow.  
Q = V×A  
Q = 2ft/sec×0.196ft2  
Q = 0.3927cfs or 0.4cfs  
22.10 Concentration  
Concentration is typically expressed as mg/l which is the weight of the constituent (mg) in 1 liter of  
water.  
As 1 liter of water weighs 1 million mg, a concentration of 1 mg/l implies 1 mg of constituent per 1  
million mg of water or one part per million (ppm). Thus, mg/l and ppm are synonymous.  
• Sometimes the constituent concentration is expressed in terms of percentage.  
Example: 12.5% chlorine concentration solution.  
100% would mean 1,000,000 mg/l or 1,000,000 ppm  
1,000,000  
=1% would be  
mg/l = 10,000 mg/l or 10,000 ppm  
100  
=12.5% chlorine concentration is 125,000 mg/l or 125,000 ppm.  
mg  
1% concentration = 10,000 ppm or  
l
mg  
0.1% concentration = 1,000 ppm or  
l
mg  
0.01% concentration = 100 ppm or  
l
mg  
10% concentration = 100,000 ppm or  
l
mg  
l
5% concentration = 50,000 ppm or  
mg  
l
12.5% concentration = 125,000 ppm or  
22.11 Density  
Density is defined as the weight of a substance per a unit of its volume. For example, pounds per  
cubic foot or pounds per gallon.  
• Here are a few key facts about density:  
Density is measured in units of lb/ft3, lb/gal, or mg/L. Density of water = 62.4 lb/ft3 = 8.34  
lb/gal.  
22.12 Specific Gravity  
• Specific gravity is the ratio of the density of a substance (liquid or solid) to the density water.  
It is the ratio of the weight of the substance of a certain volume to the weight of water of the same  
volume.  
 
 
 
358  
Chapter 22. Water Math  
Any substance with a density greater than that of water will have a specific gravity greater than 1.0.  
Any substance with a density less than that of water will have a specific gravity less than 1.0.  
• Specific gravity examples:  
Specific gravity of water = 1.0  
Specific gravity of concrete = 2.5 (depending on ingredients)  
Specific gravity of alum (liquid @ 60°F) = 1.33  
Specific gravity of hydrogen peroxide (35%) = 1.132  
• Specific gravity is used in two ways:  
1. To calculate the total weight of a % solution (either as a single gallon or a drum volume).  
Total Weight = Drum Vol X SG X 8.34  
2. To calculate the “active ingredient” weight of a single gallon or a drum.  
Active Ingredient Weight within Drum = Drum Volume X SG X 8.34 X % solution as a  
decimal. (i.e., Total Weight X % solution as a decimal)  
NOTE: Both ways start with solving for the total weight (Drum Vol X SG X 8.34). When  
solving for “active ingredient” weight, you have to then multiply by % solution as a decimal.  
Example: What is the weight of 5 gallons of a 40% ferric chloride solution given its specific gravity of  
1.43?  
(8.341.43) lbs/gal 5 gallons = 59.6 lbs  
The weight of active ferric chloride in the drum will be 59.6*0.4=23.84 lbs (as ferric chloride is 40%  
strength)  
22.13 Detention Time  
• Detention Time - The actual or theoretical (calculated) time required for water to fill a tank at a  
given flow; pass through a tank at a given flow; or remain in a settling basin, flocculating basin,  
rapid-mix chamber, or storage tank.  
Tank/clari fier volume(cu. ft or gal)  
Tank/clarifier detention time (hr) =  
Influent flow (cu. ft or gal)/hr)  
Rectangular tank/clarifier volume = width * length * depth of water  
Circular tank/clarifier volume = 0.785 * Diameter2 * depth of water  
Typically volume is calculated in cu. ft and influent flow is given in gallons. Use 7.48 gal/ft3 con-  
version factor to convert volume in cu. ft to gallons.  
22.14 Unit Conversions  
A conversion is a number that is used to multiply or divide into a measure in order to change the  
units of the original measure.  
In most instances, the conversion factor cannot be derived. It must be known. Therefore, tables  
such as the one below are used to find the common conversions.  
 
 
22.14 Unit Conversions  
359  
Measure  
Length  
Area  
Units  
inches, ft, miles  
ft2, acres  
Volume  
Density  
Flow  
ft3, gallons, acres-ft.  
weight per volume, lbs/ft3, lbs/gallon  
ft3/min, MGD, acres-ft/day  
Table 22.1: Common units in water calculations  
Some Common Conversions  
Weight  
Linear Measurements  
1 inch = 2.54 cm  
1ft3 of water = 62.4lbs  
1gal = 8.34lbs  
1foot = 30.5 cm  
1lb = 453.6grams  
1 kg = 1000 g = 2.2lbs  
1% = 10,000mg/L  
1pound = 16ozdrywt  
1ft3 = 62.4lbs  
1 meter = 100 cm = 3.281feet = 39.4 inches 1  
acre = 43,560ft2  
1yard = 3feet  
Volume  
Pressure  
1gal = 3.78 liters  
1ft = 7.48 gal  
1 L = 1000 mL  
1gal = 16cups  
Flow  
1ft of head = 0.433psi  
1psi = 2.31ft of head  
1cfs = 448gpm  
1gpm = 1440gpd  
• Common conversions in water related calculations include the following:  
gpm to cfs  
Million gallons to acre feet  
Cubic feet to acre feet  
Cubic feet of water to gallons  
gpm to MGD  
psi to feet of head  
• Steps for unit conversion:  
Step 1: Make sure the original unit is for the same measurement as the converted (desired) unit. So  
if the original unit is for area, say in ft2 the converted unit should be another area unit such as  
in2 or acre but it cannot be gallons as gallon is a unit of volume.  
Note: Calculating the weight of a certain volume of water involves the use of density which  
is the mass per volume - value in units including lbs/gallon or lbs/ft3  
360  
Chapter 22. Water Math  
Step 2: Write down the conversion formula as:  
Quantity in converted unit = Quantity (Original Unit)Conversion Factor  
Conversion unit  
(
(
(
(
(
(
(
(
(
(
Original unit  
(
(
Note: If you wish to convert cubic feet of water to pounds, you have to use its density which is the  
known mass per unit volume.  
8.34 lbs  
gallon  
62.4 lbs  
or  
ft3  
mass  
mass o f water = VolumeDensity(  
)
Volume  
Example Problems:  
1. Convert 1000 ft3 to cu. yards  
cu.yards  
1000 ft3 ∗  
= 37cu.yards  
27 ft3  
2. Convert 10 gallons/min to ft3/hr  
Note: This involves use of two conversion factors - one for converting gallons to cubic feet and  
another for converting minute to gallons.  
¨
10gallons  
ft3  
60min 80.2ft3  
¨
=
¨
min  
7.48gallons  
hr  
hr  
¨
3. Convert 100,000 ft3 to acre-ft.  
100,000ft3 ∗  
4. Convert 8 ft3 of water to pounds.  
acreft  
= 2.3acreft  
2
43,560 ft ft  
Here the conversion is from a volume (ft3) to a weight (lbs). It involves use of a standard correla-  
tion of the volume of water to its weight - its density.  
lbs  
ft3  
Weight o f water in lbs = 8ft3 62.4(  
) = 499.2 lbs  
22.15 Pounds Formula  
Pounds formula is used for:  
Calculating the quantity in pounds of a particular wastewater constituent entering or leaving a  
wastewater treatment process  
• Calculating the pounds of chemicals to be added  
So if the concentration of a particular constituent (in mg/liter) and the volume or flow of wastewater is  
given, one can calculate the amount of that constituent in pounds using the following – Pounds Formula:  
lbs  
day  
mg  
l
MG  
day  
lbs or  
= concentration( )8.34volume(MG) or flow(  
(MGD)  
Figure 22.1: Pounds formula "nomograph"  
There are three variables – (lbs, concentration and volume) and one constant (8.34) in the pounds for-  
 
22.16 Process Removal Efficiency Calculations  
361  
mula. Knowing any of the two variables in the formula, one can calculate the third (unknown) variable by  
rearranging the equation.  
22.15.1 Example Problems  
1. If the influent wastewater flow is 5 MGD and the BOD concentration is 240 mg/l what is the daily  
BOD loading in lbs/day?  
Solution:  
lbs BOD  
day  
10,000lbs  
= 5MGD240mg/l 8.34 =  
day  
2. Calculate the lbs of solids in the primary sludge if the sludge flow is 7500 gallons and the solids  
concentration is 4.5%.  
Solution  
Applying lbs formula:  
7500 MG  
1,000,000  
lbs solids =  
4.510,0008.34 = 2,815 lbs solids  
Note:  
1) 7500 gallons was converted to MG by dividing by 1,000,000  
1MG  
1,000,000 gallons  
7500 gallons∗  
2) 4.5% was converted to mg/l by multiplying by 10,000 as 1%=10,000mg/l  
3. An operator dissolves 1,200 lbs of a chemical in 12,000 gallons of water, what is the resultant  
concentration in mg/l, of the chemical solution?  
Solution:  
mg  
lbs  
Concentration  
=
=
l
Volume MG 8.34  
mg  
1,200  
11,990 mg  
Concentration  
=
or 1.2% solution  
l
0.012 8.34  
l
Note:  
1) 12,000 gallons was converted to MG by dividing by 1,000,000  
1MG  
1,000,000 gallons  
12,000 gallons∗  
22.16 Process Removal Efficiency Calculations  
• Process removal rate or removal efficiency is the percentage of the inlet concentration removed.  
It is used for quantifying the pollutant removal during wastewater treatment and is established  
based upon the amount of a particular wastewater constituent entering and leaving a treatment  
process.  
Pollutant InPollutant Out  
Process Removal Rate (%) =  
100  
Pollutant In  
If 10 units of a pollutant are entering a process and 8 units of pollutant are leaving (process re-  
moves 2 units), then the process removal rate for that pollutant is (10-8)/10*100=20%. In this  
example the process is 20% efficient in removing that particular pollutant.  
The amount of pollutant can be measured in terms of concentration (mg/l) or in terms of mass  
loading (lbs). The pounds formula is used for calculating the mass loadings.  
The above example is for calculating the removal efficiency using the inlet and outlet concentrations or  
mass loading.  
The methods below can be used for calculating either the inlet or outlet pollutant concentrations, if the  
removal efficiency and the corresponding inlet or outlet concentrations are given.  
Case 1: Calculating outlet conc. (X) given the inlet conc. and removal efficiency (RE%):  
 
 
362  
Chapter 22. Water Math  
X mg/l (Unknown)  
A mg/l (Given)  
100 mg/l  
In  
Process  
Out  
(100RE%) mg/l  
Removal E f ficiency = RE% (Given)  
Using the fact that if the inlet concentration was 100 mg/l, the outlet concentration would be 100 minus  
the removal efficiency.  
Out  
In  
X mg/l  
A mg/l  
100RE%  
Setup the equation as:  
:
=
100  
A
B
C
D
C
D
Calculate X using cross multiplication - if  
=
=A = B∗  
:
100RE%  
X mg/l = A mg/l ∗  
100  
Case 2: Calculating inlet conc. (X) given the outlet conc. and removal efficiency (RE%):  
X mg/l (Unknown)  
100 mg/l  
A mg/l (Given)  
In  
Process  
Out  
(100RE%) mg/l  
Removal E f ficiency = RE% (Given)  
Using the fact that if the inlet concentration was 100 mg/l, the outlet concentration would be 100 minus  
the removal efficiency.  
In  
Out  
X mg/l  
A mg/l  
100  
100RE%  
Setup the equation as:  
:
=
A
B
C
D
C
D
Calculate X using cross multiplication - if  
=
=A = B∗  
:
100  
X mg/l = A mg/l ∗  
100RE%  
Example Problems:  
1. What is the % removal efficiency if the influent concentration is 10 mg/L and the effluent concentra-  
tion is 2.5 mg/L?  
InOut  
102.5  
Removal Rate(%) =  
100 =⇒  
100 = 75%  
In  
10  
2. Calculate the outlet concentration if the inlet concentration is 80 mg/l and the process removal  
efficiency is 60%  
Solution:  
80mg/l  
Xmg/l  
In  
Process  
Out  
40mg/l  
100mg/l  
Removal E f ficiency = 60%  
Actual Outlet(X) 10060  
Out  
In  
:
=
80  
100  
Actual Outlet(X)  
=⇒  
= 0.4  
80  
=Actual Outlet(X) = 0.480 = 32mg/l  
22.17 Pounds Formula  
363  
3. Calculate the inlet concentration if the outlet concentration is 80 mg/l and the process removal  
efficiency is 60%  
Xmg/l  
80mg/l  
40mg/l  
In  
Process  
Out  
100mg/l  
Removal E f ficiency = 60%  
In  
Out  
Actual inlet (X)  
100  
10060  
Actual inlet (X)  
:
=
=⇒  
= 2.5  
80  
80  
Rearranging the equation: Actual inlet(X) = 2.580 = 200mg/l  
22.17 Pounds Formula  
• Pounds formula:  
lbs  
day  
mg  
l
lbs or  
= Concentration  
8.34volume(MG) or Flow(MGD)  
• So if the concentration of a particular constituent (in mg/liter) and the volume or flow of wastewater  
is given, one can calculate the amount of that constituent or using this formula.  
Important notes:  
1. The unit of the constituent loading rate will be in lbs per the unit of time the flow is expressed  
in. So if the flow is in MG per day the calculated loading rate will be in lbs/day. Likewise if the  
flow value used is in MG per minute, the calculated loading rate will be in lbs/min.  
2. If volume is used, the calculated value will be the mass of the constituent in that volume. If flow  
is used, the calculated value will be the mass of the constituent in that flow.  
3. For the Pound Formula to work, the volume or flow needs to be expressed in MG. Volume or  
flows in other units - gallons, ft3 etc. needs to be converted to MG.  
The formula assumes that all of the material found in water (TSS, BOD, MLSS, Chlorine, etc.)  
weighs the same as water, that is, 8.34 pounds per gallon.  
In the Pounds Formula, there are three variables – lbs, concentration and volume, and one constant  
- 8.34. Knowing any of the two variables in the formula, one can calculate the third (unknown)  
variable by rearranging the equation.  
Davidson Pie provides a pictorial reference for calculating any unknown variable. If for example, if  
Concentration is unknown, it can be calculated as follows:  
lbs  
day  
lbs or  
mg  
l
Concentration  
=
8.34Volume(MG) or Flow(MGD)  
• Likewise, if Volume (or Flow) is the unknown variable. it can be calculated as:  
lbs  
lbs or  
day  
Volume(MG) or Flow(MGD) =  
mg  
l
Concentration  
8.34  
 
364  
Chapter 22. Water Math  
lbs or lbs/day  
÷
=
Concentration  
mg/l  
Volume(MG)  
Flow(MGD)  
8.34  
X
X
Multiply  
Multiply  
Figure 22.2: Davidson Pie  
• Pounds formula is used for:  
Calculating the quantity in pounds of a particular wastewater constituent entering or leaving a  
wastewater treatment process  
Calculating the pounds of chemicals to be added  
Example 1: If a 5 MGD flow is to be dosed with 25 mg/l of a certain chemical, calculate the lbs/day that  
chemical required.  
Solution  
Applying lbs formula:  
lbs  
day  
mg  
l
lbs  
= 5MGD250  
8.34 = 1,042  
day  
Example 2: Calculate the lbs of chemical in 7,500 gallons of 4.5% active solution of that chemical.  
Solution  
Applying lbs formula:  
7500  
lbschemical =  
MG4.510,0008.34 = 2,815 lbs chemical  
1,000,000  
Note:  
1) 7500 gallons was converted to MG by dividing by 1,000,000  
1MG  
1,000,000 gallon  
7500 gallons∗  
2) 4.5% was converted to mg/l by multiplying by 10,000 as 1%=10,000mg/l  
22.18 Chemicals Related Math Problems  
22.18.1 Chemical Dosing  
• Use lbs formula to calculate the lbs of chemicals required  
Using the calculated lbs chemical required value, calculate the amount of that chemical at the  
concentration available  
So for example, if asked how much many gallons per day of bleach solution (SG 1.2)containing 12.5%  
available chlorine is required to disinfect a 10 MGD flow of water given the required chlorine dosage of 7  
mg/l.  
 
 
22.18 Chemicals Related Math Problems  
365  
1. calculate the lbs of chlorine required using the lbs formula:  
mg  
l
=10MGD 7  
8.34 = 583.8 lbs chlorine per day  
2. calculate the gallons of bleach which will provide the 583.8 lbs chlorine  
Applying the lbs formula - note that 8.34 * SG will give the actual lbs/gal of bleach. If SG is not  
provided, use only 8.34 lbs per gallon:  
lbs bleach  
day  
gal  
day  
lbs bleach lbs chlorine  
0.0125  
gal lb bleach  
583.8  
= x  
8.341.2  
gal  
day  
583.8  
8.341.20.125  
gal  
day  
=x  
=
= 467  
The above problem can be solved directly using the formula below given in the SWRCB Water  
Treatment Exam Formula Sheet.  
(MGD)(ppm or mg/l)8.34 lbs/gal  
% purityChemical Wt. (lbs/gal)  
1078.34  
gal  
day  
GPD =  
GPD =  
= 467  
0.125(1.28.34)  
22.18.2 Chemical Dilution  
Chemicals obtained in bulk are typically delivered in higher concentrations to reduce transportation  
costs and need dilution to ensure proper mixing and dosage control.  
• Thus, for dilution calculations, if:  
C1 and V1 is the concentration and volume respectively of the concentrated chemical used for the  
dilution, and  
C2 and V2 is the concentration and volume of the product after dilution with water  
As, the mass of the target chemical in the volume of the concentrated product used for dilution will  
remain the same in the final diluted product:  
C1 * V1 = C2 * V2.  
• Thus, knowing C1, C2 and V2, we can calculate V1 as:  
C2 V2  
V1 =  
C1  
Example Problem:  
How much initial volume of a 4% polymer solution is needed to make 3500 gallons of polymer at 0.25%  
concentration.  
Solution:  
C.25% V.25%  
0.25 3500  
V4%  
=
=
= 219gal  
C4%  
4
So take 219 gallons of the 4% polymer and dilute to 3500 gallons to give a 0.25% polymer solution.  
 
366  
Chapter 22. Water Math  
22.19 Preliminary Treatment  
22.20 Preliminary Treatment Math Problems  
Preliminary Treatment math problems relate to the following:  
22.20.1 Channel Velocity and Flow Rate  
Flow Rate - Q (volume/time) = velocity (distance or length traveled /time) * surface area  
Velocity is the speed at which the water is flowing. It is measured in units of length/time – ft./sec.  
Velocity of water flowing through can be calculated by dividing the flow rate by area of the flow stream.  
volume or cubic length  
time  
sur face area in the direction of flowsquare length  
flow rate(  
)
length  
time  
Velocity  
=
For a flow in a channel:  
Example Problems:  
1. Calculate the velocity of a 14 MGD flow in a 6 ft wide channel with a water depth of two feet.  
Flow(Q) = Velocity(V)Area(A)  
MG 106gal  
ft3  
day  
ft  
sec  
ft3  
sec  
ft3  
sec  
=14  
= V  
6 ft 2 ft =21.7  
= 12V  
day  
ft  
MG  
21.7  
12  
7.48gal 246060  
ft  
=V  
=
= 1.8  
sec  
sec  
2. Calculate the flow, in gpd, that would pass through a grit chamber 2 feet wide, at a depth of 6  
inches, with a velocity of 1 ft /sec  
Solution:  
Q = V A  
ft  
ft3  
s
Q = 1 (20.5)ft2 = 1  
s
ft3 (144060)s  
gal  
ft3  
gal  
day  
£
Q = 1  
7.48  
= 646,272  
s
day  
£
3. A wastewater channel is 3.25 feet wide and is conveying a wastewater flow of 3.5 MGD. The  
wastewater flow is 8 inches deep. Calculate the velocity of this flow.  
 
 
 
22.21 Primary Treatment  
367  
Solution:  
Q
Q = V A =V =  
A
3
¨
¨
day  
¨
£
ft  
MG 1000000gal  
¨
3.5  
7.48gal  
¨
¨
ft  
s
ft  
s
day  
MG  
(144060)s  
¨
¨
=V  
=
= 2.2  
(3.250.75) ft2  
4. A plastic float is dropped into a wastewater channel and is found to travel 10 feet in 4.2 seconds.  
The channel is 2.4 feet wide and is flowing 1.8 feet deep. Calculate the flow rate of this wastewater  
in cubic feet per second.  
Solution:  
Q = V A  
ft3  
s
10 ft  
4.2s  
ft3  
s
=Q  
=
(2.41.8) ft2 = 10.3  
22.20.2 Grit Removal Rates  
Typical grit removal ranges from 0.5 to 30 ft3/MG  
Example Problems:  
1. At a wastewater treatment plant which receives a flow rate of 650,000 gallons per day, a total of 50  
cubic feet of grit was removed for the month. Calculate the rate of grit removal assuming 30 days  
in a month.  
Solution:  
ft3  
MG  
ft3  
month  
gal  
ft3  
MG  
¨
day  
¨
GritRemoval  
= 50  
1,000,000  
= 2.6  
¨
month 30days  
MG  
650,000gal  
¨
22.21 Primary Treatment  
22.21.1 Hydraulic or Surface Loading Rate  
The hydraulic or surface loading rate measures how rapidly wastewater moves through the primary clar-  
ifier. It is measured in terms of the number of gallons flowing each day through one square foot surface  
area of the clarifier.  
gpd  
ft2  
Clari fier in fluent flow(gpd)  
Clari fier sur face area( ft2)  
Clarifier hydraulic loading  
=
Rectangular clarifier surface area = width * length  
Circular clarifier surface area = 0.785 * Diameter2  
22.21.2 Detention Time  
Detention time is the length of time that wastewater stays in the settling tank is called the detention time.  
It is also the time it takes for a unit volume of wastewater to pass entirely through a primary clarifier  
 
 
 
 
368  
Chapter 22. Water Math  
Clari fier volume(cu. ft or gal)  
Clarifier detention time (hr) =  
Influent flow (cu. ft or gal)/hr)  
Rectangular clarifier volume = width * length * depth of water  
Circular clarifier volume = 0.785 * Diameter2 * depth of water  
Typically volume is calculated in cu. ft and influent flow is given in gallons. Use 7.48 gal/ft3 conversion  
factor to convert volume in cu. ft to gallons.  
22.21.3 Weir Overflow Rate  
The weirs at the end of the primary clarifier allow for the even distribution of the the outlet flow across  
the entire length of the weir. An adequate length of weir is needed to ensure smooth and even flow of  
wastewater over the weirs. Weir overflow rate measures the number of gallons of wastewater per day  
flowing over one foot of weir.  
gpd  
ft  
Clari fier in fluent flow(gpd)  
Total e f fluent weir length ( ft)  
Weir over flow rate  
=
Circular clarifier weir length = 3.14 * Diameter  
Example problem for (a), (b) and (c) above:  
A circular clarifier receives a flow of 11 MGD. If the clarifier is 90 ft. in diameter and is 12 ft. deep, what  
is: a) the hydraulic/surface loading rate, b) clarifier detention time in hours, and c) weir overflow rate?  
a) Hydraulic/surface loading rate:  
11MG 106gal  
¨
¨
¨
gpd  
ft2  
day  
0.785902 ft2  
MG  
¨
Clarifier hydraulic loading  
==  
= 1,730gpd/ft2  
b) Clarifier detention time:  
Clarifier volume(cu. ft or gal)  
In fluent flow (cu. ft or gal)/hr)  
Clarifier detention time (hr) =  
(0.785902 12)ft3  
Clarifier detention time (hr) =  
= 1.2hrs  
6
ft3  
¨
¨
day  
¨
11MG 10 gal  
¨
24hrs  
7.48gal  
¨
¨
day  
¨
MG  
¨
c) Weir overflow rate:  
11MG 106gal  
¨
¨
¨
gpd  
ft  
day  
3.1490 ft  
MG  
¨
Weir over flow rate  
=
= 38,924gpd/ ft  
22.21.4 Removal Efficiency  
Primary sedimentation removes suspended wastewater solids which includes BOD. The efficiency of  
the primary is established as the percentage of the amount of parameter removed. The parameter may  
quantified as mass (lbs) or as concentration (mg/l).  
Parameter InParameter Out  
Removal e f ficiency(%) =  
100  
Parameter In  
For TSS removal:  
 
 
22.21 Primary Treatment  
369  
TSSIn (mg/l)TSSOut (mg/l)  
TSSIn (mg/l)  
TSS Removal e f ficiency(%) =  
For BOD removal:  
BOD Removal e f ficiency(%) =  
22.21.5 Solids Removal  
100  
BODIn (mg/l)BODOut (mg/l)  
BODIn (mg/l)  
100  
Type 1 Problems: These involve calculating lbs of solids removed given any two of the following TSS  
parameters - inlet concentration, outlet concentration and removal efficiency.  
a. If the inlet and outlet concentrations are given, calculate the mg/l of TSS removed using:  
TSSremoved = TSSin(mg/l)TSSout(mg/l)  
Then knowing the flow, use the lbs formula to calculate the lbs solids removed.  
b. If either inlet or outlet concentration is given along with the clarifier removal efficiency, using the  
removal efficiency calculate the unknown outlet concentration (if only the inlet is given) or the inlet  
concentration (if only the outlet is given)  
i) If inlet and removal efficiency is given, calculate the outlet by subtracting the product of inlet and  
removal efficiency from the inlet.  
TSSout = TSSin (TSSin %Removal)  
Example if the removal efficiency is 60% and the inlet concentration is 300mg/l:  
TSSout = 3003000.6 = 120mg/l  
ii) If outlet and removal efficiency is given, calculate the inlet concentration by dividing the outlet by  
(1-removal efficiency).  
TSSout  
TSSin =  
1%Removal  
Example if the removal efficiency is 60% and the outlet concentration is 120mg/l:  
120  
10.6  
TSSin =  
= 300mg/l  
TSSin TSSout  
Note: You may derive the above formulas by algebraically manipulating: %Removal =  
TSSin  
Example Problem:  
How many lbs of solids are removed daily by a primary clarifier treating a 6 MGD flow if the average  
influent TSS concentration is 300 mg/l and the clarifier TSS removal efficiency is 67%.  
TSSout = (300mg/l 3000.67) = 99mg/l  
lbs solids removed = (30099)mg/l 8.346MGD = 10,058 lbs solids removed per day  
Type 2 Problems: These involve calculating the amount of sludge pumping given the solids removed. The  
solids removed from the primary clarifier is sludge with a typical solids concentration of about 3% to 5%.  
 
370  
Chapter 22. Water Math  
Given the amount of total solids removed and given the sludge concentration, the volume of sludge pump-  
ing can be calculated as follows:  
ft3 sludge pumped  
day  
lbs solids (removed)  
1 lb sludge  
gal sludge  
ft3 sludge  
=
∗ ∗  
(%) lbs solids 8.34lb sludge 7.48 gal  
day  
So for the solids removed in the above example, if the primary sludge has 5% solids, the required sludge  
pumping can be calculated as:  
(
(
(
(
(
ft3 sludge 10,058 lbs solids  
1 lb sludge  
ga(l sludge  
ft3 sludge  
ft3 sludge  
day  
(
(
(
(
=
∗ ∗ ∗  
(
0.05 lbs solids 8.34lb sludge  
(
= 3,224  
(
(
(
day  
day  
7.48 gal  
22.22 Trickling Filter Calculations  
Trickling filter problems involve calculation of the following:  
22.22.1 Hydraulic or surface loading  
• Hydraulic or surface loading is expressed as gpd/ ft2  
• The gpd is the total flow (QT )to the filter - primary influent flow + recirculated flow(QT = QI + QR)  
Example Problem:  
The total influent flow (including recirculation) to a trickling filter is 1.89 MGD. If the trickling filter is 80  
ft in diameter, what is the hydraulic loading in gpd/sq ft on the trickling filter?  
Solution:  
gpd  
ft2  
(1.89106)gpd  
(0.785802)ft2  
gpd  
ft2  
Hydraulic loading  
=
= 376  
22.22.2 BOD and TSS Removal  
BOD and removal is based upon the TF influent and effluent concentrations.  
Inconc Outconc  
%Removal =  
100  
Inconc  
Example Problem:  
The suspended solids concentration entering a trickling filter is 236 mg/l. If the suspended solids con-  
centration of the trickling filter effluent is 33 mg/l, what is the suspended solids removal efficiency of the  
trickling filter?  
Solution:  
236mg/l 33mg/l  
%Removal =  
100 = 86%  
236mg/l  
22.22.3 Organic Loading  
• Organic loading to a trickling filter is typically expressed as lbs BOD/(day-1000 cu ft).  
• The lbs/day BOD value is the BOD loading from the primary effluent.  
• The 1000 ct. ft is the volume of the media.  
• The media volume is calculated by multiplying the TF surface area by the media height.  
As the dimensions are typically given in ft., calculate the volume in ft3 and then divide the calcu-  
lated volume by 1000 to give the volume in units of 1000ft3  
 
 
 
 
22.23 Stabilization Pond Calculations  
371  
Example Problem:  
A trickling filter, 70 ft in diameter with a media depth of 6 ft, receives a flow of 0.78 MGD. If the BOD  
concentration of the primary effluent is 167 mg/L, what is the organic loading on the trickling filter in lbs  
BOD/day/1000 cu ft?  
lbs BOD  
lbs BOD feed to TF per day  
Solution: Organic loading :  
=
day1000ft3  
volume in 1000ft3  
(0.781678.34)lbs BOD  
47lbs BOD  
day  
=
=
1000 ft3  
(0.785702 6) ft3 ∗  
1000 ft3  
day1000 ft3  
22.22.4 Recirculation Ratio  
Recirculated Flow(QR)  
In fluent Flow(QI)  
Recirculation Ratio(RR) =  
Total Flow(QT )Influent Flow(QI)  
In fluent Flow(QI)  
Recirculation Ratio(RR) =  
Total Flow(QT ) = In fluent Flow(QI)(Recirculation Ratio(RR)+1) Make sure QR  
are the same in a given problem  
, QT and QI units  
Example Problems:  
1. The influent to the trickling filter is 1.61 MGD. If the recirculated flow is 2.27 MGD, what is the  
recirculation ratio?  
QR  
QI  
2.27  
1.61  
Solution: RR =  
=
= 1.4  
2. A trickling filter has a total flow of 32 MGD. If the recirculation ratio is 0.8, what is the primary  
effluent flow to the TF?  
Solution:  
Total Flow(QT ) = In fluent Flow(QI)(Recirculation Ratio(RR)+1)  
32  
1.8  
=32MGD = QI (0.8+1) =QI =  
= 17.8MGD  
22.23 Stabilization Pond Calculations  
22.23.1 Pond Area  
Formula: Pond Area = WidthLength  
Pond Volume  
Pond Depth  
also, Pond Area =  
Example Problem:  
A pond is 260 ft. long and 80 ft. wide. What is the area of this pond in acres?  
Solution:  
acre  
43,560ft2  
(26080) ft2 ∗  
= 0.48acre  
22.23.2 Solids Loading Rate  
lbs TSS  
day  
Formula: Pond TSS loading rate =  
Example Problem:  
 
 
 
 
372  
Chapter 22. Water Math  
The influent flow to a pond is 10,000 gallons/hour with a suspended solids concentration of 142mg/L in  
the raw wastewater. How many lbs of suspended solids are sent to the pond daily?  
Solution:  
lbs TSS  
day  
gal 24hrs  
hr  
MG  
mg  
l
lbs TSS  
8.34 = 284  
day  
= 10,000  
142  
day 1,000,000gal  
22.23.3 Organic Loading Rate  
Formula: Pond organic loading rate =  
Example Problem:  
lbs BOD/day  
Area(acre)  
The flow to a pond is 7.2MGD. If the pond diameter is 350 ft and the BOD in the pond influent is 170mg/L,  
what is the organic loading to this pond in lbs BOD/day/acre?  
Solution:  
lbs BOD per day (7.2MGD 170mg/l 8.34) 43,560ft2  
4,624lbs BOD  
dayacre  
Organic loading =  
=
=
area (acres)  
0.7853502 ft2  
acre  
22.23.4 Detention time  
Volume  
Flow  
Formula: Pond detention time =  
Example Problem:  
A 40 acre wastewater treatment pond receives a flow of 0.6 MGD. If the pond is operated at a depth of 4ft.  
What is the detention time of this pond?  
Solution:  
Volume  
Flow  
(404)acreft  
ft3  
Pond detention time =  
=
= 87 days  
gal  
acreft  
0.6106  
day 7.48gal 43,560ft3  
22.23.5 Hydraulic Loading Rate  
"
#
in  
day  
Flow  
Area  
Formula:Pond hydraulic loading rate  
=
"
#
in  
day  
Pond depth (in)  
also, Pond hydraulic loading rate  
=
Volume  
Flow  
Pond detention time  
The second formula above is because:  
Flow  
Hydraulic Loading (HL) =  
Area  
Vol  
Flow  
Vol  
DT  
Detention time (DT) =  
=Flow =  
Substituting for flow in the HL formula above:  
Vol  
DT  
Area  
Vol  
AreaDT  
Pond Depth  
Vol  
Area  
HL =  
or  
=HL =  
as  
= Pond Depth  
DT  
Example Problems:  
 
 
 
22.24 Activated Sludge Calculations  
373  
1. Find hydraulic loading in inches/day for a pond given the following:  
• Pond depth = 12ft.  
• Pond volume = 1,400,000ft3  
• Pond flow = 1,000,000gal/day  
Solution:  
"
#
in  
day  
Flow  
Area  
Pond hydraulic loading rate  
ft3  
=
gal  
1,000,000  
in  
12 = 13.8  
ft  
in  
day 7.48gal  
1,400,000 ft3  
12 ft  
=⇒  
day  
Note: The area of the pond was found by dividing the volume (1,000,000 ft3) by the pond depth (12ft)  
2. Find pond hydraulic loading in inches/day when the depth of the pond is 6 ft. and the detention  
time is 30 days.  
Solution:  
Pond depth (in)  
Volume  
Pond detention time  
Flow  
Pond hydraulic loading rate =  
612 inches  
30 days  
2.4in  
day  
=⇒  
=
22.24 Activated Sludge Calculations  
22.24.1 Mean Cell Residence Time  
The MCRT is calculated as:  
MCRT(days) =  
Total MLSS lbs in the aeration system (aeration tank + clari fier)  
Total amount in lbs/day o f suspended solids leaving the system (E f fluent SS+WAS solids)  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
E f fluent suspended solids (lbs/day)+ WAS SS (lbs/day)  
Key Points for Solving MCRT Problems  
1. MLSS quantification:  
Pounds formula is used to calculate lbs MLSS using: i) aeration tank and the clarifier vol-  
umes, and ii) the given MLSS concentration.  
The MLSS concentrations for the aeration tank and the clarifier are the same. So the given  
MLSS concentration applies to both - the aeration tank and the clarifier  
• Make sure it is the MLSS concentration that you are using not the MLVSS concentration.  
If MLSS concentration is not given but instead MLVSS concentration is given, you will need  
to find the MLSS concentration by dividing the MLVSS conc. by the mixed liquor volatile  
solids, as MLVSS(conc.) = MLSS * % volatile solids  
2. Suspended solids quantification:  
(a) Effluent suspended solids  
• Effluent suspended solids can be quantified using the pounds formula - using the effluent  
flow (in MGD) and the effluent suspended solids concentration.  
 
 
374  
Chapter 22. Water Math  
(b) WAS suspended solids  
Use pounds formula given the WAS flow (make sure it is in MG) and the WAS SS  
concentration  
Note that the WAS and RAS streams have the same SS concentration. If WAS SS con-  
centration is not specified, use the RAS SS concentration  
Example Problems:  
1. In an conventional activated sludge plant the aeration tank contains 6000 lbs of MLSS and the final  
clarifier contains 2300 lbs of MLSS. 1450 lbs of solids are wasted each day and 90 lbs/day of solids  
leave in the final effluent. Calculate the MCRT for this plant.  
Solution:  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
6000lbs + 2300lbs  
90lbs/day + 1450lbs/day  
MCRT(days) =  
= 5.4 = 5days  
2. A activated sludge plant treats an average influent flow of 4 MGD. The plant has two aeration  
tanks – 0.45 MG volume each and two final clarifiers – 0.2 MG volume each, and a mixed liquor  
suspended solids concentration averages 1800 mg/l. The effluent suspended solids concentration  
averages 18 mg/L. The WAS flow is 100,000 gallons per day has a SS concentration of 6100 mg/L.  
Calculate the MCRT  
Solution:  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
MLSS in aeration tank (lbs) = 20.4518008.34 = 13511lbs  
MLSS in clarifier (lbs) = 20.218008.34 = 6005lbs  
SS e f fluent (lbs/day) = 4MGD18mg/l 8.34 = 600lbs/day  
100000  
1000000  
SS WAS (lbs/day) =  
MGD4800mg/l 8.34 = 4003lbs/day  
13511+6005  
Plugging in the values calculated above: MCRT(days) =  
= 4.2 = 4days  
600+4003  
22.24.2 F:M (Food to Microorganism Ratio  
This parameter ratios the food – the mass of primary effluent BOD entering the aeration basin to  
the mass of the microorganisms - MLVSS, in the aeration basin.  
Only the mass of the microorganisms (MLVSS) in the aeration basin is used – the mass of  
microorganisms in the secondary clarifier is not considered  
• Common ranges for F/M for a conventional activated sludge plant are from 0.15 to 0.5.  
• The optimum F/M varies from plant to plant and can be determined by trial and error.  
The F:M may be used to determine the concentration of mixed liquor suspended solids to be main-  
tained in the aeration tank.  
Generally, low F/M ratios should be carried during the colder months as the microorganism activity  
(metabolism) is lower.  
F:M and MCRT are inversely related: that is a long MCRT means a low F:M and a short MCRT  
means a high F:M  
amount o f food coming in  
amount o f microorganisms present  
F:M=  
 
22.24 Activated Sludge Calculations  
375  
(lbs/day) primary e f fluent BOD entering the aeration tank  
=
(lbs) MLVSS in the aeration tank  
Key Points for Solving F:M Problem  
1. Quantifying F:  
• use the pounds formula to calculate the lbs/day of BOD in the primary effluent.  
lbs/day BOD = Primary eff. flow (MGD)* Primary eff. BOD concentration (mg/l) * 8.34  
2. Quantifying M:  
The concentration of the microorganisms is assumed to be the same as the MLVSS concentra-  
tion  
Only the mass of the microorganisms (MLVSS) in the aeration basin is used – the mass  
of microorganisms in the secondary clarifier is not considered  
lbs MLVSS may be calculated using pounds formula using the volume of the aeration tank  
(in MG) and the MLVSS concentration  
If the MLVSS concentration is not given, it can be calculated from the MLSS and MLSS %  
volatile matter (solids) concentration  
MLVSS = MLSS * % MLSS volatile solids  
Example Problem:  
i. A conventional activated sludge plant receives an average flow of 5.5 MGD. The influent  
BOD to the plant averages 230mg/l and the primary effluent BOD average 160 mg/l. The 1  
MG aeration tank has an MLSS concentration of 2800 mg/L and the MLVSS volatile solids  
content is 75%. Calculated the F:M ratio for this plant. Solution:  
F=5.5*160*8.34=7339lbs/day BOD  
M=1*2800*0.75*8.34=17514lbs MLVSS  
7339  
17514  
F:M=  
= 0.41  
Note: The 160 mg/l BOD concentration of the primary effluent was used for the F calculation  
and not 230mg/l - which is the BOD concentration of the flow coming into the plant  
22.24.3 Sludge Volume Index (SVI)  
• SVI measures the settleability and compactibility of the secondary sludge  
• It is calculated using results from the 30-minute settleability test and the MLSS concentration  
SVI is expressed in ml/g and it is essentially the volume (ml) of 1 gram of the MLSS after 30  
minutes of settling  
it provides a more accurate picture of the sludge settling characteristics than settleability or  
MLSS alone  
50 to 120 ml/gm SVI value is considered optimal. Higher SVI values indicate sludge that  
is slow to settle and not compacting well. When SVI values are approaching 200 ml/gm,  
activated sludge process is considered to be "bulking".  
Settled sludge volume in ml/l after 30 min  
MLSS mg/l  
mg  
g
SVI (ml/g)=  
1000  
Key Points for Solving SVI Problems  
For the settling test MLSS is typically settled in a 1 liter settleometer. The volume of the settled  
solids is therefore read as ml/L. So if for any reason a larger or smaller volume of the mixed liquor  
sample is taken, the settle solids value should commensurate with the volume of the MLSS sample.  
For example, if a 2 liter settleometer is used and if the solids settle to 400 ml in that settleometer,  
 
376  
Chapter 22. Water Math  
the ml/L will be 400ml/2L or 200ml/L  
For some problems, the settled solids volume is provided as a percentage (%). So if a 1-liter sett-  
lometer is used and the settled solids volume is reported as 25%, it implies a settled sludge volume  
of 250ml/L  
Example Problem:  
i. In an aeration tank, the MLSS is 2500 mg/l and recorded 30-minute settling test indicates 230 ml/L.  
What is the sludge volume index?  
Solution:  
230ml/l  
mg  
g
SVI=  
1000  
= 92ml/g  
2500mg/l  
22.24.4 Sample Math Problems  
1. An activated sludge plant operates well at an F:M ratio of between 0.23 and 0.28. Calculate the  
minimum MLSS concentration, given the following:  
Q = 0.4 MGD  
Primary influent BOD = 250 mg/l  
Primary effluent BOD = 128 mg/l  
Aeration tank vol. = 350,000 gallons  
Clarifier vol = 250,000 gallons  
MLSS has 80% volatile solids  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
1
=F : M ∝  
=F:M is inversely proportional to MLSS  
MLSS concentration  
So to have minimum MLSS conc. F:M needs to be the maximum of the range provided  
0.41288.34  
If the MLSS concentration = x: =F : M = 0.28 =  
=x = 5,446mg/l MLSS  
0.35x0.8  
 
22.25 Solids Thickening Calculations  
377  
2. Given that an activated sludge plant with an influent flow of 1.2 MGD is operated at an MCRT of 6  
days and the parameters below, calculate the WAS flow rate (wasting rate) in gallon per day.  
Solution:  
Two aeration tanks – 0.5 MG each  
Two final clarifiers – 0.25 MG each  
WAS – 7500 ppm  
mg  
l
Final effluent = 20  
mg  
MLSS –3600  
L
MLSS volatile solids content = 80%  
lbs MLSS(system)  
MCRT =  
lbs  
day  
lbs  
day  
E f fluentSS  
+
WASSS  
Step 1: Calculate the lbs MLSS (system):  
lbs MLSS (system) = (20.5+20.25)MG3,600  
Step 2: Calculate the lbs/day EffluentSS:  
mg  
L
8.34 = 45,036 lbs  
lbs  
day  
mg  
L
E f fluentSS = 1.2MG20  
8.34 = 200.2lbs  
Step 3: Plug in the values in the MCRT formula:  
45,036  
MCRT: 6 days =  
lbs  
lbs  
200.2  
+
WASSS  
day day  
Step 4: Solving for WASSS:  
lbs  
day  
45,036  
lbs  
day  
WASSS  
=
200.2 = 7,306  
6
Step 5: Solving for WAS flow using the lbs formula:  
lbs  
day  
7,306  
= WAS Flow (MGD)7,5008.34  
7,306  
gal  
day  
=WAS Flow (MGD) =  
= 0.116MGD = 116,000  
7,5008.34  
22.25 Solids Thickening Calculations  
1. Calculate the air required (SCFM) to meet a 0.04 lb air:lb feed solids ratio for a 100 GPM WAS  
flow with a solids content of 6500mg/l? Assume 0.08 lbs air/SCF air. Solution:  
lb air  
lb solids  
0.08 lbs air SCF X SCF per minute  
= 0.04 =  
100  
MG  
mg  
l
6,500  
8.34 lbs solids  
1,000,000 min  
0.045.421  
0.08  
=x SCF per minute =  
= 2.7 SCFM  
2. A treatment plant receives an influent flow of 30 MGD with a TSS concentration of 280 mg/l. The  
primary treatment removes 55% TS and the primary sludge is pumped to a 40 ft diameter gravity  
 
378  
Chapter 22. Water Math  
thickener. Calculate the average solids loading to the thickener in lbs TSS/day-ft2  
Solution:  
(30 MGD 2800.55 mg/l 8.34)lbs TSS per day  
0.785402 ft2  
Solids loading to gravity thickener=  
=
30.7 lbs TSS/dayft2  
22.26 Digester Calculations  
22.26.1 Sludge Pumping  
Example Problems:  
1. A primary clarifier receives an average flow of 12 MGD containing 280mg/L of TSS. This clarifier  
typically removes 75% TSS and produces a 3.5% sludge and the sludge pump is rated to pump 35  
cu.ft/min.  
a. How many pounds of TSS is removed in the clarifier?  
b. How many cu.ft of sludge at the given 3.5% sludge needs to be pumped per day to remove  
the solids?  
c. How many minutes would the sludge pump need to be operational each day to pump the  
required amount of sludge - calculated from ii. above?  
d. For how many minutes each hour the sludge pump should be programmed to operate (Given  
the number of minutes the pump need to operate per day calculated from iii. above) ?  
Solution:  
a. lbs solids removed = (2800.75)mg/l 12MGD8.34 = 21,017 lbs solids per day  
(
21,017 lbs solids x ft3 sludge 7.48 ga(l sludge 0.035 lbs solids 8.34 lb sludge  
(
(
(
(
b.  
=
(
ft3 sludge  
(
ga(l sludge  
(
day  
day  
1 lb sludge  
x ft3 sludge  
day  
21,017  
0.0358347.48  
ft3 sludge  
=
= 9,626  
day  
9,626 ft3 sludge  
min  
275 minutes  
c.  
d.  
=
day  
35 ft3 sludge  
day  
275 minutes  
day  
24 hrs  
11.4 minutes  
=
day  
hr  
2. Calculate the lbs/day of solids removed in a primary clarifier treating a 5 MGD flow with an aver-  
age influent and effluent TSS concentrations of 250 mg/l and 98 mg/l respectively. Solution:  
lbs  
day  
mg  
l
lbs  
= 5MGD(25098)  
8.34 = 6338  
day  
22.26.2 Sludge Blending  
In digestion the feed is typically a blend of primary and secondary sludges - each of which have  
different total and organic solids content.  
• Thus, for blending calculations, if:  
C1 and V1 is the concentration and volume (flow) respectively of the primary sludge portion of the  
digester feed, and  
 
 
 
22.26 Digester Calculations  
379  
C2 and V2 is the concentration and volume (flow) respectively of the secondary sludge feed C3 and  
and V3 is the concentration and volume (flow) respectively of the digester sludge feed  
The sum of the mass from each of the two source sludge feeds will equal to the mass in the digester  
sludge feed:  
C1 * V1 + C2 * V2 = C3 * V3.  
This equation can be manipulated algebraically to calculate anyone of the unknown values in the  
equation.  
Also, any of the three volume variables can be expressed as the sum or difference of the other two -  
, or V1 + V2 = V3 or V1 = V3 - V2 or V2 = VT - V1  
Example Problem: Two sludges are blended together as follows: 15,000 gal. primary sludge at 4.1%  
solids. 28,000 gal. secondary sludge at 1.3% solids. What is the combined solids concentration? If the  
primary sludge is 68% VS and the secondary sludge is 63% VS, what is the VS concentration (%) in the  
combined sludge?  
Solution:  
Combined solids concentration:  
C1 V1 +C2 V2 = C3 V3  
C1 V1 +C2 V2 C1 V1 +C2 V2  
4.115,000+1.328,000  
15,000+28,000  
=C3 =  
=
=
= 2.28%  
V3  
V1 +V2  
Lbs of VS in combined sludge:  
CVS1 V1 +CVS2 V2 = CVS3 V3  
CVS1 V1 +CVS2 V2 CVS1 V1 +CVS1 V2  
=CVS3  
=
=
V3  
V1 +V2  
4.115,0000.68+1.328,0000.63  
15,000+28,000  
=CVS3  
=
= 1.50%  
22.26.3 Digester VS or Organic Loading  
Example Problems:  
1. How many pounds of TS and VS are pumped to a digester each day if the digester receives 10,000  
gpd of sludge at 5% solids concentration with an average VS% of 75%?  
Solution:  
Digester TS loading (lbs/day)  
lbs TS 10,000 gal sludge (8.340.05)lbs TS)  
lbs TS  
day  
=
= 4,170  
day  
day  
gal sludge  
Digester VS loading (lbs/day)  
lbs TS  
day  
lbs VS  
lb TS  
lbs VS  
day  
= 4,170  
0.75  
= 3,128  
2. An anaerobic digester is 37’ in diameter and 27’ deep with a 5,000 gallon daily sludge flow. The  
sludge is 6% solids and 66% volatile solids. What is the volatile solids loading in pounds per cubic  
foot per day?  
 
380  
Chapter 22. Water Math  
lbs VS  
Solution:  
Digester Loading  
day  
Digester volatile solids loading rate =  
Digester volume(V) ft3  
gal sludge  
lbs VS  
gal sludge  
5000  
(8.340.060.66)  
lbs VS  
dayft3  
day  
=
= 0.057  
π
(
372 27)ft3  
4
22.26.4 Total volatile solids (VS) reduction  
This provide a measure of the organic matter content removed and converted into digester gas in  
the digester.  
• Higher volatile solids reduction implies higher gas production and lower biosolids hauling costs.  
• The VS reduction of the digester is provided by the Van Kleeck equation  
VSin VSout  
Digester VS reduction(%) =  
100  
VSin VSin VSout  
Digester volatile solids concentration is typically expressed as a percentage of the sludge total  
solids.  
• 70% VS which means that 70% of the total solids is volatile solids.  
The value of VSin and VSout for the digester VS reduction (Van Kleek) equation above should be in  
fraction and not as a percentage.  
A value of 0.7 should be used in the equation if the VS concentration is 70%. Likewise, as 0.525 if  
the VS concentration is 52.5%.  
Applying this equation to calculate the digester VS reduction if the inlet sludge VS averages 75%  
and the outlet sludge is 58%?  
Example Problem:  
Calculate the % VS reduction in a digester given the volatile solids content of the influent sludge to  
the digester is 70% and the volatile solids content of the sludge leaving the digester is 52.5%  
0.70.525  
Solution: Digester VS reduction(%) =  
100 = 53%  
0.70.70.525  
22.27 Dewatering math problems  
Solving Dewatering Problems  
1. Calculation of solids recovery  
(a) Calculate amount of solids fed to the dewatering unit  
(b) Calculate the amount of solids produced as part of the dewatered cake  
(c) The ratio of the solids in dewatered cake to that in the feed times 100 will give you the solids  
recovery (solids recovery rate)  
2. Calculation of dewatered cake volume  
 
 
22.27 Dewatering math problems  
381  
(a) First calculate the amount of cake solids produced in terms of weight per time.  
(b) From the weight of the cake produced calculate the volume - from the cake density which is  
normally given  
3. Calculation of solids hauling costs or savings associated with change in cake solids content  
(a) First calculate the amount of dry solids produced as part of the original wet cake solids  
percent  
(b) Using the value of the dry solids calculate the wet cake weight with the new cake solids  
percentage  
4. General formula for calculating net savings associated with change in cake solids content  
(New solids %Old Solids %)  
Savings =  
Old Cost  
New solids %  
So if the cake dryness goes up from 20% to 26% and currently a utility is spending $ 1,000,000 per  
year for biosolids hauling and disposal, their net savings will be: (26%-%20)/26*$ 1,000,000 = $  
230,769  
22.27.1 Calculation of solids recovery  
Example Problems:  
(a) Calculate amount of solids fed to the dewatering unit  
(b) Calculate the amount of solids produced as part of the dewatered cake  
(c) The ratio of the solids in dewatered cake to that in the feed times 100 will give you the solids  
recovery (solids recovery rate)  
22.27.2 Calculation of dewatered cake volume  
(a) First calculate the amount of cake solids produced in terms of weight per time.  
(b) From the weight of the cake produced calculate the volume - from the cake density which is normally  
given  
22.27.3 Hauling cost impact due to solids content change  
(a) First calculate the amount of dry solids produced as part of the original wet cake solids percent  
(b) Using the value of the dry solids calculate the wet cake weight with the new cake solids percentage  
General formula for calculating net savings associated with change in cake solids content:  
(New solids(%)Old Solids(%)  
Savings =  
Old Cost  
Old solids(%)  
So if the average cake dryness goes up from 20% to 26% and currently this utility is spending $ 1,000,000  
per year for biosolids hauling and disposal, their net savings will be:  
(26%20%)  
1,000,000 = $300,000  
20%  
Example Problems  
1. 12,000 ft3 of anaerobically digested sludge containing 2.8% TS is dewatered in a centrifuge. The  
centrifuge yields 37 yd3 of 26% of dewatered cake with a density of 73 lb/ft3. Calculate the solids  
capture rate.  
Solution:  
gal (8.340.028lbs TS)  
= 20,960lbs TS  
lbs TS feed to centrifuge = 12,000ft3 sludge7.48  
ft3  
gal sludge  
ft3 (73lbs0.26 TS)  
lbs TS feed from centrifuge = 37yd3 sludge 27  
= 18,961lbs TS  
yd3  
ft3 sludge  
 
 
 
382  
Chapter 22. Water Math  
18,961lbs solids produced by centri fuge  
20,960lbs solids fed from digester  
solids capture rate =  
100 = 90.4%solids capture  
2. At a 60 GPM of 2.8% feed a belt press which has a 90% solids capture rate produces a 20% cake at  
68 lbs/ ft3. How long would it take to fill a 3 yd3 bin  
Solution:  
cake TS produced lbs 60gallons sludge 8.34lbs sludge feed 0.028lbs TS  
=
0.9  
lb sludge  
min  
min  
galllon  
12.61lbs TS  
=
min  
ft3 cake produced  
min  
12.61lbs TS 100lbs cake  
ft3 cake  
68lbs cake  
0.927 ft3cake  
=
=
=
=
min  
20lbs TS  
min  
ft3 cake produced  
min  
12.61lbs TS 100lbs cake  
ft3 cake  
68lbs cake  
0.927 ft3cake  
min  
20lbs TS  
min  
min  
0.927 ft3  
27 ft3  
yd3  
Time required to fill the bin =  
3yd3 ∗  
= 75min  
Chapter Assessment  
Practice Problems - Fractions  
1
1. Convert 22 into a fraction  
4
2. Express 10ft, 6in into fraction  
3. Express 10ft, 6in into decimal  
Practice Problems - Decimals and Powers of Ten  
1. Write the equivalent of 10,000,000 as a power of ten  
2. Find the product of 3.4564102  
3. Find the product of 534.567102  
165.93  
4. Find the value of  
102  
5. Find the value of 0.023104  
Solutions:  
1. 107  
2. 345.64  
3. 5.34567  
4. 16,593  
5. 230  
384  
Chapter 22. Water Math  
Practice Problems - Rounding and Significant Digits  
Round the following to the nearest hundredths (the second place after the decimal).  
A. 2.4568  
B. 27.2534  
C. 128.2111  
D. 364.8762  
E. 354.777777  
F. 34.666666  
G. 67.33333  
Solution:  
A. 2.46  
B. 27.25  
C. 128.21  
D. 364.88  
E. 354.78  
F. 34.67  
G. 67.33  
Round the following to the nearest tenths (the first place after the decimal).  
A. 2.4568  
B. 27.2534  
C. 128.2111  
D. 364.8762  
E. 354.777777  
F. 34.666666  
G. 67.33333 Solution:  
A. 2.5  
B. 27.3  
C. 128.2  
D. 364.9  
E. 354.8  
F. 34.7  
G. 67.3  
Round the following answers off to the most significant digit.  
Problem  
Accurate Answer  
A.  
B.  
C.  
D.  
25.1+26.43 = 51.53  
128.456121.4 = 7.056  
857.92432 = 77.07568  
8.564+5 = 13.564  
22.27 Dewatering math problems  
385  
Problem  
Accurate Answer  
A.  
B.  
C.  
D.  
26.34×124.34567 = 3,275.26495  
23.58×34.251 = 807.63858  
12,453/13.9 = 895.8992805755  
12,457.92×3 = 37,373.76  
Practice Problems - Averages  
1. Find the average of the following set of numbers:  
0.2  
0.2  
0.1  
0.3  
0.2  
0.4  
0.6  
0.1  
0.3  
2. The chemical used for each day during a week is given below. Based on these data, what was the  
average lb/day chemical used during the week?  
Monday  
Tuesday  
Wednesday  
Thursday  
Friday  
92 lb/day  
93 lb/day  
98 lb/day  
93 lb/day  
89 lb/day  
93 lb/day  
97 lb/day  
Saturday  
Sunday  
3. The average chemical use at a plant is 77 lb/day. If the chemical inventory is 2800 lbs, how many  
days supply is this?  
4. A well pumped for 45 days. The beginning meter reading was 7,456,400 and 45 days later the same  
meter was 15,154,400. What was the average flow in gallons per day?  
Practice Problems - Percentage  
1. 25% of the chlorine in a 30 -gallon vat has been used. How many gallons are remaining in the vat?  
2. The annual public works budget is $147,450. If 75% of the budget should be spent by the end of  
September, how many dollars are to be spent? How many dollars will be remaining?  
3. A 75 pound container of calcium hypochlorite has a purity of 67%. What is the total weight of the  
calcium hypochlorite?  
386  
Chapter 22. Water Math  
4. 3/4 is the same as what percentage?  
5. A 2% chlorine solution is what concentration in mg/L ?  
6. A water plant produces 84,000 gallons per day. 7,560 gallons are used to backwash the filter. What  
percentage of water is used to backwash?  
7. The average day winter demand of a community is 14,500 gallons. If the summer demand is esti-  
mated to be 72% greater than the winter, what is the estimated summer demand? Demand - When  
related to use, the amount of water used in a period of time. The term is in reference to the "de-  
mand" put onto the system to meet the need of customers.  
8. The master meter for a system shows a monthly total of 700,000 gallons. Of the total water, 600,000  
gallons were used for billing. Another 30,000 gallons were used for flushing. On top of that,  
15,000 gallons were used in a fire episode and an estimated 20,000 gallons were lost to a main  
break that was repaired that same day. What is the total unaccounted for water loss percentage for  
the month?  
9. Your water system takes 75 coliform tests per month. This month there were 6 positive samples.  
What is the percentage of samples which tested positive?  
Total volume to be pumped  
Time =  
Pump flow rate  
7.48gal  
(0.7851102 25) ft3 ∗  
ft3  
=⇒  
= 1,251 min  
1420gal  
min  
Practice Problems - Ratio and Proportion  
1. It takes 6 gallons of chlorine solution to obtain a proper residual when the flow is 45,000 gpd. How  
many gallons will it take when the flow is 62,000 gpd?  
2. A motor is rated at 41 amps average draw per leg at 30Hp. What is the actual Hp when the draw is  
36 amps? C.  
3. If it takes 2 operators 4.5 days to clean an aeration basin, how long will it take three operators to do  
the same job?  
4. It takes 3 hours to clean 400 feet of collection system using a sewer ball. How long will it take to  
clean 250 feet?  
5. It takes 14 cups of HTH to make a 12% solution, and each cup holds 300 grams. How many cups  
will it take to make a 5% solution?  
6. A bike travelling at 5 miles/hr completes a journey in 40 minutes. How long would the same jour-  
ney take if the speed was increased to 8 miles/hr?  
Solution  
1. The gallons chlorine and flow are directly related.  
Thus,  
6
X
662,000  
45,000  
=
=X =  
= 8.3gallons  
45,000 62,000  
22.27 Dewatering math problems  
387  
2. The amp draw and Hp are directly related.  
This  
30  
41 36  
X
3036  
=
=X =  
= 26.3Hp  
41  
3. The number of operators and the days to clean are inversely related.  
Thus,  
24.5  
24.5 = 3X =X =  
= 3days  
3
4. The hours to clean and the length of system cleaned are directly proportional.  
Thus,  
3
X
3250  
=
=X =  
= 1.9hours  
400 250  
400  
5. The cups of HTH and percentage HTH solution are directly proportional.  
Thus,  
14  
12  
X
5
145  
=
=X =  
= 5.8cups  
12  
6. The bike speed and time to complete the journey are inversely related.  
Thus,  
540  
540 = 8X =X =  
= 25min  
8
Practice Problems - Area and Volume  
1. A 60-foot diameter tank contains 422,000 gallons of water. Calculate the height of water in the  
storage tank.  
2. What is the volume of water in ft3, of a sedimentation basin that is 22 feet long, and 15 feet wide,  
and filled to 10 feet?  
3. What is the volume in ft3 of an elevated clear well that is 17.5 feet in diameter, and filled to 14 feet?  
4. What is the area of the top of a storage tank that is 75 feet in diameter?  
5. What is the area of a wall 175ft. in length and 20ft. wide?  
6. You are tasked with filling an area with rock near some of your equipment. 1 Bag of rock covers  
250 square feet. The area that needs rock cover is 400 feet in length and 30 feet wide. How many  
bags do you need to purchase?  
7. A circular clearwell is 150 feet in diameter and 40 feet tall. The Clearwell has an overflow at 35  
feet. What is the maximum amount of water the clearwell can hold in Million gallons rounded to  
the nearest hundredth?  
8. A sedimentation basin is 400 feet length, 50 feet in width, and 15 feet deep. What is the volume  
expressed in cubic feet?  
9. A clearwell holds 314,000ft3 of water. It is 100ft in diameter. What is the height of the clearwell?  
10. A treatment plant operator must fill a clearwell with 10,000ft3 of water in 90 minutes. What is the  
rate of flow expressed in GPM?  
11. A water tank has a capacity of 6MG. It is currently half full. It will take 6 hours to fill. What is the  
flow rate of the pump?  
12. A clearwell with the capacity of 2.5MG is being filled after a maintenance period. The flow rate is  
388  
Chapter 22. Water Math  
2,500 GPM. The operator begins filling at 7 AM. At what time will the clearwell be full?  
13. A chemical feed pump with a 6-inch bore and a 6-inch stroke pumps 60 cycles per minute. Find the  
pumping rate in gpm.  
14. Determine the flow capacity of a pump in gpm if the pump lowers the water level in a 6 -foot  
square wet well by 8 inches in 5 minutes.  
15. How much paint will it take for a single coat of the top and sidewalls of the storage tank that is  
100-feet in diameter and 30-feet tall, if one gallon of paint covers 200 square feet?  
a. 86 gallons  
b. 96 gallons  
c. 106 gallons  
d. 116 gallons  
e. 126 gallons  
16. Under like conditions, how much more water would an 8-inch pipe carry than a 4-inch pipe?  
a. 2 times  
b. 3 times  
c. 4 times  
d. not enough information given  
Solution:  
Volume  
1. Volume = Surface area * height =height =  
Surface area  
Diameter=60’  
Volume=422,000 gal  
Height=h’  
ft  
3
ft  
422,000 gal∗  
7.48 gal  
=height =  
= 101 ft  
0.785602 ft2  
Practice Problems - Flow and Velocity  
1. A rectangular channel 3 ft. wide contains water 2 ft. deep flowing at a velocity of 1.5 fps. What is  
the flow rate in cfs?  
2. Flow in an 8-inch pipe is 500 gpm. What is the average velocity in ft/sec? (Assume pipe is flowing  
full)  
3. A pipeline is 18” in diameter and flowing at a velocity of 125 ft. per minute. What is the flow in  
gallons per minute?  
4. The velocity in a pipeline is 2 ft./sec. and the flow is 3,000 gpm. What is the diameter of the pipe in  
inches?  
5. Find the flow in a 4-inch pipe when the velocity is 1.5 feet per second.  
6. A 42-inch diameter pipe transfers 35 cubic feet of water per second. Find the velocity in ft/sec.  
7. A plastic float is dropped into a channel and is found to travel 10 feet in 4.2 seconds. The channel  
22.27 Dewatering math problems  
389  
is 2.4 feet wide and 1.8 feet deep. Calculate the flow rate of water in cfs.  
8. The flow velocity of a 6-inch diameter pipe is twice that of a 12-inch diameter pipe if both are  
carrying 50 gpm of water. True or false?  
9. What should the flow meter read in gpm if a 4-inch diameter main is to be flushed at a velocity of  
4.6 fps?  
10. The velocity through a channel is 4.18 fps. If the channel is 4 feet wide by 2 feet deep by 10 feet  
long, what is the flow rate in gpm?  
11. What is the average flow velocity in ft/sec for a 12-inch diameter main carrying a daily flow of  
2.5mgd ?  
Solution:  
1. Solution:  
ft  
sec  
ft3  
sec  
Q = V A =Q = 1.5  
(32)ft2 = 9  
2. Solution:  
Q = V A  
ft  
min  
3
ft ∗  
¨
min  
¨
500gallon  
60sec  
Q
A
ft  
s
7.48gal  
=V =  
=V  
=
= 3.2 ft/s  
2
8
12  
2
0.785∗  
ft  
3. Solution:  
The diameter of the pipe is 4 inches. Therefore, the radius is 2 inches. Convert the 2 inches to feet.  
2
12  
= 0.6667ft  
A = π ×r2  
A = π ×(0.167ft)2  
A = π ×0.028ft2  
A = 0.09ft2  
Q = V×A  
Q = 1.5ft/sec×0.09ft2  
Q = 0.14ft/3sec(cfs)  
390  
Chapter 22. Water Math  
Practice Problems - Unit Conversions  
Convert the following:  
1. Convert 1000 ft3 to cu. yards  
2. Convert 10 gallons/min to ft3/hr  
3. Convert 100,000 ft3 to acre-ft.  
4. Find the flow in gpm when the total flow for the day is 65,000 gpd.  
5. Find the flow in gpm when the flow is 1.3cfs.  
6. Find the flow in gpm when the flow is 0.25cfs.  
7. The flow rate through a filter is 4.25 MGD. What is this flow rate expressed as gpm?  
8. After calibrating a chemical feed pump, you’ve determined that the maximum feed rate is 178 mL/  
minute. If this pump ran continuously, how many gallons will it pump in a full day?  
9. A plant produces 2,000 cubic foot of water per hour. How many gallons of water is produced in an  
8-hour shift?  
Solution  
1. Solution:  
cu.yards  
27 ft3  
1000 ft3 ∗  
= 37cu.yards  
2. Solution:  
¨
10 gallons  
ft3  
60min 80.2 ft3  
¨
=
¨
min  
7.48gallons  
hr  
hr  
¨
3. Solution:  
acreft  
100,000ft3 ∗  
= 2.3acreft  
2
43,560 ft ft  
Note: From the conversion table: acre = 43,560 ft2  
Thus, acre-ft = 43,560 ft2-ft or 43,560 ft3  
4. Solution:  
65,000gpd  
1,440 min/day  
= 45gpm  
5. Solution:  
cfs 448gpm  
1.3  
x
= 582gpm  
1
1cfs  
6. Solution:  
cfs 448gpm  
0.25  
×
= 112gpm  
1
1cfs  
7. Solution:  
Flowrate,gpm =  
Flow rate, gpd  
1440 min/day  
Note: We are assuming that the filter operated uniformly over that 24 hour period.  
¨
MG  
gal  
MG  
¨
¨
4.25  
1,000,000  
¨
¨
day  
¨
Flowrate,gpm =  
8. Solution:  
= 2,951 gpm  
min  
1440  
¨
day  
¨
22.27 Dewatering math problems  
391  
2000 ft3 7.48 gallons 60 hr  
119,680 gallons  
=
ft3  
hr  
shift  
shift  
Practice Problems - Concentration  
1. What is the concentration in mg/l of 4.5% solution of that substance.  
2. How many lbs of salt is needed to make 5 gallons of a 2,500mg/l solution  
Solution  
1. 45,000 mg/l  
2. Applying pounds formula: lbs salt =  
5
1,000,000  
2,5008.34 = 0.14lbs  
Practice Problems - Density and Specific Gravity  
1. What is the specific gravity of a 1 ft3 concrete block which weighs 145 lbs?  
2. What is the specific gravity of a chlorine solution if 1 (one) gallon weighs 10.2lbs?  
3. How much does each gallon of zinc orthophosphate weigh (pounds) if it has a specific gravity of  
1.46?  
4. How much does a 55 gallon drum of 25% caustic soda weigh (pounds) if the specific gravity is  
1.28?  
Practice Problems - Detention Time  
1. A flocculation basin is 7 ft deep, 15 ft wide, and 30 ft long. If the flow through the basin is 1.35  
MGD, what is the detention time in minutes?  
2. A tank has a diameter of 60 feet with an overflow depth at 44 feet. The current water level is 16  
feet. Water is flowing into the tank at a rate of 250 gallons per minute. At this rate, how many days  
will it take to fill the tank to the overflow?  
3. How long will it take to fill a 50 gallon hypochlorite tank if the flow is 5gpm ?  
4. Find the detention time in a 45,000 gallon reservoir if the flow rate is 85gpm.  
5. If the fuel consumption to the boiler is 35 gallons per day. How many days will the 500 gallon tank  
last.  
6. The sedimentation basin in a water plant contains 5,775 gallons. What is the detention time if the  
flow is 175gpm.  
Solution  
1. Solution:  
392  
Chapter 22. Water Math  
7’ deep  
1.5 MGD  
15’ wide  
30’ long  
gal  
ft3  
(30157)ft3 7.48  
DT =  
= 25min  
gal  
day  
1,350,000  
day 1440min  
2. Solution:  
250 gpm  
Diameter=60’  
Overflow level = 44’  
Current level = 16’  
7.48gallons  
0.785602 (4416) ft3 ∗  
ft3  
Volume  
Flow  
Fill time =  
=
= 1.6 days  
gallons 1440 min  
250  
min  
day  
3. Solution:  
50gal  
5gal/min  
4. Solution:  
45,000gal  
DT =  
= 10 min  
529 min  
60 min/hr  
DT =  
= 529 min or  
= 14.3 days  
= 33 min  
= 8.8hrs  
85gal/min  
5. Solution:  
500 gal  
DT =  
35gal/ day  
6. Solution:  
5,775gal  
DT =  
175gal/min  
22.27 Dewatering math problems  
393  
Practice Problems - Pounds Formula  
1. A water treatment plant operates at the rate of 75 gallons per minute. They dose soda ash at 14  
mg/L. How many pounds of soda ash will they use in a day?  
2. A water treatment plant is producing 1.5 million gallons per day of potable water, and uses 38  
pounds of soda ash for pH adjustment. What is the dose of soda ash at that plant?  
3. A water treatment plant produces 150,000 gallons of water every day. It uses an average of 2  
pounds of permanganate for iron and manganese removal. What is the dose of the permanganate?  
4. A water treatment plant uses 8 pounds of chlorine daily and the dose is 17 mg/l. How many gallons  
are they producing?  
5. An operator mixes 40 lb of lime in a 100-gal tank containing 80 gal of water. What is the percent of  
lime in the slurry?  
6. A treatment plant has a maximum output of 30MGD and doses ferric chloride at 75 mg/L. How  
many pounds of Ferric Chloride does the plant use in a day?  
7. A treatment plant uses 750 pounds of alum a day as it treats 15MGD. What was the dose rate?  
8. A treatment plant operates at 1,500 gallons a minute and uses 500 pounds of alum a day. What is  
the alum dose?  
Solution:  
1. Solution:  
lbs/day  
14 mg/l  
75 GPM  
8.34  
lbs  
day  
MG  
day  
mg  
l
= Flow  
Concentration  
8.34  
MG  
¨
gallons  
lbs  
day  
min  
mg  
lbs  
¨
= 75  
1440  
250  
8.34 = 225  
l day  
¨
min  
day 1,000,000 gallons  
¨
2. Solution:  
38 lbs/day  
?
mg/l  
1.5 MGD  
8.34  
394  
Chapter 22. Water Math  
lbs  
lbs  
day  
MG  
day  
mg  
l
mg  
l
mg  
l
day  
MG  
day  
= Flow  
Concentration  
8.34 =Concentration  
=
Concentration  
=
Flow  
8.34  
lbs  
38  
mg  
l
day  
= 3  
MG  
1.5  
8.34  
day  
3. Solution:  
38 lbs/day  
?
mg/l  
1.5 MGD  
8.34  
lbs  
lbs  
day  
MG  
day  
mg  
l
mg  
l
mg  
l
day  
MG  
day  
= Flow  
Concentration  
8.34 =Concentration  
=
Concentration  
=
Flow  
8.34  
lbs  
2
mg  
l
day  
 
!
= 3  
Gallons  
MG  
150,000  
8.34  
(
(
(
(
day  
1,000,000 Gallons  
4. Solution:  
8 lbs/day  
17 mg/l  
?
MGD  
8.34  
lbs  
day  
MG  
day  
mg  
l
= Flow  
Concentration  
8.34  
lbs  
day  
lbs  
day  
8
MG  
day  
MG  
day  
=Flow  
=
=
= 0.056425  
mg  
l
mg  
l
Concentration  
8.34 17  
8.34  
MG 1,000,000 Gallons  
0.056425  
= 56,425 Gallons  
day  
MG  
5. Solution:  
mg  
l
lbs = Volume(MG)Concentration  
8.34  
22.27 Dewatering math problems  
395  
38 lbs/day  
?
mg/l  
1.5 MGD  
8.34  
mg  
l
lbs  
40 lbs  
MG  
1,000,000 gallons  
=Concentration  
=
=
Volume(MG)8.34  
80 gallons∗  
8.34  
Practice Problems - Temperature Conversion  
1. Convert 22C into degree Fahrenheit.  
2. Convert 56C into degree Celsius.  
Preliminary Treatment  
1. On an average a 12.5 yd. load of grit is hauled to the landfill once every 20 days. Plant flow aver-  
ages 12.5 MGD. Calculate the rate of grit collection in ft3/MG.  
Correct Answer(s):  
a. 1.4  
2. A grit channel is 2 feet 4 inches wide. When wastewater is flowing 8 inches deep in this channel,  
the velocity is found to be 1.6 ft per second. Calculate flow in MGD.  
a. 1.55 MGD  
*b. 1.60 MGD  
c. 1.90 MGD  
d. 2.50 MGD  
3. A grit channel is 2 feet 4 inches wide. When wastewater is flowing 8 inches deep in this channel,  
the velocity is found to be 1.6 ft per second. Calculate flow in MGD.  
a. 1.55 MGD  
*b. 1.60 MGD  
c. 1.90 MGD  
d. 2.50 MGD  
4. At a wastewater treatment plant which receives a flow rate of 650,000 gallons per day, a total of 50  
cubic feet of grit was removed for the month. Calculate the rate of grit removal assuming 30 days  
in a month.  
396  
Chapter 22. Water Math  
Solution:  
GritRemoval MG = 50  
ft3  
MG  
3
3
¨
ft  
ft  
d¨ay  
gal  
month  
¨
650,000gal 1,000,000MG = 2.6  
¨
month  
30days  
5. On an average a 12.5 yd. load of grit is hauled to the landfill once every 20 days. Plant flow aver-  
ages 12.5 MGD. Calculate the rate of grit collection in ft3/MG.  
Solution:  
ft3  
MG  
3
3
3
ft  
MG  
27 ft  
d¨ay  
12.5yd  
¨
GritRemoval  
=
= 1.4  
3
¨
20days  
12.5MG  
yd  
¨
6. On an average, 2 inches of grit is collected and removed every day in a 2.2 feet wide, 205 feet long  
grit channel. Knowing the average flow through that grit channel is 10 MGD calculate the rate of  
grit collection in ft3/MG  
Solution:  
3
75.16 ft  
day  
day  
ft3  
MG  
3
75.16ft  
day  
2
Grit volume accumulated: 12 ft (2.2205) ft2 =  
Grit Collection:  
= 7.5  
10MG  
7. What is the grit production rate (cu. ft/MG) if 46 cu. ft of grit is removed every five days and the  
daily plant flow averages 2 MGD.  
a. 12 cu. ft/MG  
b. 23 cu. ft/MG  
*c. 4.6 cu. ft/MG  
d. 9.2 cu. ft/MG  
Solution:  
ft3  
MG  
3
3
ft  
MG  
46ft  
d¨ay  
¨
Grit Production Rate  
=
= 4.6  
days 2MG  
5
8. A grit chamber is 2 feet 4 inches wide. When wastewater is flowing 8 inches deep in this channel,  
the flow velocity is found to be 1.6 ft per second. Calculate flow in MGD.  
Solution:  
Q = V A  
Q = 1.6  
ft  
s
28  
12 12  
8
ft3  
s
ft2 = 2.49  
ft3 (144060)s  
gal MG  
£
Q = 2.49  
7.48  
= 1.61MGD  
s
day  
£
ft3  
gal  
9. A steel bin is used to collect the daily accumulation of grit for disposal in a land fill. The bin is 8  
ft long, 5.25 ft. wide, and 6 feet deep. When the bin is three-quarters full it is hauled away. This  
occurs on the average every 10 days. Average flow to the plant is  
3.75 MGD. What is the rate of grit collection in units of cubic feet per MG of flow?  
Correct Answer(s):  
22.27 Dewatering math problems  
397  
a. 4.9  
10. A flow controlled grit chamber has two channels which are each 2.5 feet wide, and 20 feet long.  
Only one of these channels is currently in service and it receiving a flow of 4.5 MGD. The wastewa-  
ter is flowing 1.5 feet deep in this channel. Calculate the velocity of this  
flow. Should we put the other channel in service?  
Correct Answer(s):  
11. Calculate the flow, in gpd, that would pass through a grit chamber 2 feet wide, at a depth of 6  
inches, with a velocity of 1 ft /sec.  
Correct Answer(s):  
a. 646272.0  
Primary Treatment  
1. A clarifier has a TSS removal efficiency of 50%. If the influent TSS concentration is 220 mg/L,  
how many lbs/day of TSS are removed if the flow is 10 MGD. Also, how many cu. ft of sludge is  
pumped if the sludge has a TS concentration of 5%.  
lbs solids removed = (2200.50)mg/l 10MGD8.34 = 9,174lbs solids per day  
(
(
(
(
(
ft3 sludge 9,174 lbs solids  
1 lb sludge  
ga(l sludge  
ft3 sludge  
ft3 sludge  
day  
(
(
(
(
=
∗ ∗ ∗  
(
0.05 lbs solids 8.34lb sludge  
(
= 2,941  
(
(
(
day  
day  
7.48 gal  
2. A community has a total flow of 15 MGD which is passed through a primary treatment plant which  
removes 60% of the TSS and 35% of the BOD. The average strength of the influent is 400 mg/l  
TSS and 275 mg/l BOD. If the total solids of the raw sludge is 5%, how many cu. ft of sludge is  
pumped daily? lbs solids removed = (4000.60)mg/l15MGD8.34 = 30,024lbs solids per day  
(
(
(
(
(
ft3 sludge 30,924 lbs solids  
1 lb sludge  
ga(l sludge  
ft3 sludge  
ft3 sludge  
day  
(
(
(
(
=
∗ ∗ ∗  
(
0.05 lbs solids 8.34lb sludge  
(
= 9,626  
(
(
(
day  
day  
7.48 gal  
3. How many lbs of solids are removed daily by a primary clarifier treating a 6 MGD flow if the  
average influent TSS concentration is 300 mg/l and the clarifier TSS removal efficiency is 67%?  
As the removal efficiency is 67%, 0.67 * 300 mg/l = 201 mg/l solids are removed.  
The total lbs removed can be calculated using the lbs formula.  
lbs solids  
mg SS  
l
lbs solids  
day  
= 6MGD201  
8.34 = 10,058  
day  
4. Calculate the primary clarifier influent solids concentration if its outlet concentration is 60 mg/l and  
the known clarifier removal efficiency is 75%?  
Actual inlet (X)  
Actual outlet  
Actual inlet (X)  
60  
100  
=
100Removal e f ficiency  
100  
= 4  
=
10075  
=Actual inlet (X) = 460 = 240mg/l  
5. A circular clarifier receives a flow of 11 MGD. If the clarifier is 90 ft. in diameter and is 12 ft. deep,  
what is: a) the hydraulic/surface loading rate, b) weir overflow rate, and c) clarifier detention time  
in hours?  
Solution:  
398  
Chapter 22. Water Math  
a) Hydraulic/surface loading rate:  
6
10 gal  
11MG  
day  
gpd  
ft  
= 1,730gpd/ ft2  
2  
MG  
Clarifier hydraulic loading  
==  
2
2
0.78590 ft  
b) Weir overflow rate:  
6
10 gal  
11MG  
day  
gpd  
ft  
MG  
3.1490ft  
Weir over flow rate  
=
= 38,924gpd/ft  
c) Clarifier detention time:  
Clarifier volume(cu. ft or gal)  
In fluent flow (cu. ft or gal)/hr)  
Clarifier detention time (hr) =  
2
3
(0.78590 15)ft  
Clarifier detention time (hr) =  
= 2hrs  
day  
3
6
gal  
ft  
24hrs  
7.48gal  
10 ꢀ  
11MG  
day  
MG  
6. A clarifier has a TSS removal efficiency of 50%. If the influent TSS concentration is 220 mg/L,  
how many lbs/day of TSS are removed if the flow is 10 MGD. Also, how many cu. ft of sludge is  
pumped if the sludge has a TS concentration of 5%.  
Solution:  
lbs solids removed = (2200.50)mg/l 10MGD8.34 = 9,174lbs solids per day  
(
(
(
(
(
ft3 sludge 9,174 lbs solids  
1 lb sludge  
ga(l sludge  
ft3 sludge  
ft3 sludge  
day  
(
(
(
(
=
∗ ∗ ∗  
(
0.05 lbs solids 8.34lb sludge  
(
= 2,941  
(
(
(
day  
day  
7.48 gal  
7. If a clarifier has a capacity of 0.25 MG, what is the detention time in hours if it receives a flow of 3  
MGD  
Solution:  
Clarifier volume(MG)  
Clarifier detention time (hr) =  
In fluent flow (MG/hr)  
¨
0.25MG  
¨
Clarifier detention time (hr) =  
= 2hrs  
day  
3MG  
ꢀ ꢀ  
24hrs  
day  
8. If a clarifier has a capacity of 0.25 MG, what is the detention time in hours if it receives a flow of 3  
MGD  
Solution:  
Clarifier volume(MG)  
Clarifier detention time (hr) =  
In fluent flow (MG/hr)  
¨
0.25MG  
¨
Clarifier detention time (hr) =  
= 2hrs  
day  
3MG  
ꢀ ꢀ  
24hrs  
day  
9. At a 2.5 MGD wastewater treatment plant the primary clarifier has a detention time of 2 hours.  
How many gallons does this clarifier hold?  
Solution:  
Clarifier volume(gal)  
Clarifier detention time (hr) =  
In fluent flow (gal/hr)  
=Clarifier volume(gal) = Clari fier detention time (hr)Influent flow (gal/hr)  
   
gal  
day  
=Clarifier volume(gal) = 2 hrs 2.5106  
= 208,333 gals  
day 24 hrs  
10. What is the surface loading rate (gal/(day-sq.ft) of a 15 MGD flow in a 105 ft diameter primary  
sedimentation tank operating at water depth of 20 ft.  
Solution:  
22.27 Dewatering math problems  
399  
15MG 106gal  
¨
¨
¨
gpd  
ft2  
day  
MG  
¨
Clarifier hydraulic loading  
=
= 1,733gpd/ft2  
0.7851052 ft2  
11. If a 90 ft diameter primary clarifier operating at water depth of 20 ft is treating a 12MGD flow,  
calculate the surface loading rate (gal/(day-sq.ft).  
Solution:  
12MG 106gal  
¨
¨
¨
gpd  
ft2  
day  
0.785902 ft2  
MG  
¨
Clarifier hydraulic loading  
=
= 1,887gpd/ft2  
12. What is the weir overflow rate (gpd/ft) when treating a 15 MGD flow in a 105 ft diameter primary  
sedimentation tank operating at water depth of 20 ft.  
Solution:  
(0.7851052 20) ft3  
Clarifier detention time (hr) =  
= 2.1hrs  
6
ft3  
¨
¨
day  
¨
15MG 10 gal  
¨
¨
24hrs  
7.48gal  
¨
¨
day  
MG  
¨
13. A circular clarifier receives a flow of 3.2 MGD. If the diameter is 70 ft, what is the weir overflow  
rate?  
Solution:  
3.2MG 106gal  
¨
¨
¨
gpd  
ft  
day  
3.1470 ft  
MG  
¨
Weir over flow rate  
=
= 14,559gpd/ ft  
14. How many lbs/day of solids are removed in a clarifier treating a 6 MGD flow if the average inlet  
concentration is 320 mg/l and its average outlet concentration is 80 mg/l  
Solution:  
Applying pounds formula:  
lbs solids  
= 6MGD(32080)mg 8.34 = 120,096  
lbs solids removed  
day  
l
day  
15. At a 2.5 MGD wastewater treatment plant the primary clarifier has a detention time of 2 hours.  
How many gallons does this clarifier hold?  
16. What is the detention time (hrs) in a 105’ diameter primary sedimentation tank operating at water  
depth of 20’ when treating a 15 MGD flow (Ans: 2.1)  
17. What is the surface loading rate (gal/(day-sq.ft) of a 15 MGD flow in a 105 ft diameter primary  
sedimentation tank operating at water depth of 20 ft. (Ans: 1,733)  
18. What is the weir overflow rate (gpd/ft) when treating a 15 MGD flow in a 105 ft diameter primary  
sedimentation tank operating at water depth of 20 ft. (Ans: 45,496)  
19. A circular clarifier receives a flow of 3.2 MGD. If the diameter is 70 ft, what is the weir overflow  
rate? (Ans: 14,559)  
20. A clarifier has an influent suspended solids concentration of 420 mg/L. If the suspended solids  
efficiency is 62%, how many mg/L suspended solids are removed? (Ans: 260)  
21. Given the following for a primary sedimentation tank:  
TSS removal efficiency = 63%  
Effluent TSS concentration = 95 mg/L  
400  
Chapter 22. Water Math  
Calculate the influent TSS concentration (mg/L)  
(Ans: 257)  
22. What is the clarifier influent TSS if its outlet concentration is 60 mg/l and the known clarifier  
removal efficiency is 75%?  
*a. 240 mg/l  
b. 300 mg/l  
c. 80 mg/l  
d. 120 mg/l  
23. What is the clarifier removal efficiency if the inlet and outlet concentrations are 300 mg/l and 60  
mg/l respectively?  
a. 20%  
b. 30%  
*c. 80%  
d. 72%  
24. A rectangular sedimentation tank is 85 feet long, 35 feet wide, and 14 feet deep including 3 feet of  
freeboard. Flow to this tank is 2.3 MGD. Calculate the surface loading to this tank in gpd per ft2.  
a. 318 gpd/ft2  
*b. 773 gpd/ft2  
c. 845 gpd/ft2  
d. 1932 gpd/ft2  
25. What is the detention time in a 40 ft diameter primary clarifier operating with a 12 ft water depth,  
when treating a 1.5 MGD flow?  
a. 0.24 hours  
b. 0.60 hours  
*c. 1.8 hours  
d. 2.3 hours  
e. 11.0 hours  
26. How many lbs/day of solids are removed in a clarifier treating a 6 MGD flow if the average inlet  
concentration is 320 mg/l and its average outlet concentration is 80 mg/l  
a. 16,000 lbs/day  
*b. 1,210 lbs/day  
c. 4,000 lbs/day  
d. 9,000 lbs/day  
27. At a 5 MGD wastewater flow, a primary clarifier has a detention time of 1 hour. How many gallons  
does this clarifier hold?  
a. 104,000 gallons  
*b. 208,000 gallons  
c. 250,000 gallons  
d. 500,000 gallons  
e. 5,000,000 gallons  
28. Percent efficiency of total solids or BOD removal is calculated using the following formula: (In-  
Out*100)/(In-(In*Out))  
a. True  
*b. False  
29. The only function of flights in a rectangular sedimentation tank is to push the settled sludge towards  
22.27 Dewatering math problems  
401  
the hopper  
a. True  
*b. False  
30. A settling basin has a length of 60 ft., width of 20 ft., and depth of 12 ft. At a flow rate of 60 mgd,  
what is the retention time of this basin?  
a. 15 minutes.  
b. 30 minutes.  
*c. 1.1 hours.  
d. 2.3 hours.  
31. Calculate the detention time for a sedimentation tank that is 48 feet wide, 210 feet long and 9 feet  
deep with a flow of 5 MGD.  
*a. 3.25 hours.  
b. 3.63 hours.  
c. 5.65 hours.  
d. 5.82 hours.  
32. Calculate the weir loading for a sedimentation tank that has an outlet weir 480 ft. long and a flow  
of 5 MGD.  
a. 9220 gpd/ft.  
b. 9600 gpd/ft.  
c. 9920 gpd/ft.  
*d. 10,420 gpd/ft.  
33. If a primary clarifier consistently operates at 30 % efficiency and produces an effluent which aver-  
ages 140 mg/l BOD, what is the influent BOD?  
a. 100 mg/l  
b. 125 mg/l  
c. 175 mg/l  
*d. 200 mg/l  
e. 225 mg/l  
34. At a 2.5 MGD wastewater treatment plant the primary clarifier has a detention time of 2 hours.  
How many gallons does this clarifier hold?  
a. 104,000 gallons  
*b. 208,000 gallons  
c. 250,000 gallons  
d. 500,000 gallons  
e. 5,000,000 gallons  
35. A 3.1 MGD flow with a 190 mg/l TSS concentration is treated in a primary clarifier which averages  
55% removal efficiency. Calculate the pounds/day TSS in the primary effluent.  
a. 2445 lbs/day  
*b. 2211 lbs/day  
c. 3800 lbs/day  
d. 4200 lbs/day  
36. A 75 ft long, 25 ft wide and 10 ft deep sedimentation basin receives a flow of 1.6 MGD. What is  
the detention time in hours.  
a. 7 hrs  
b. 1.7 hrs  
402  
Chapter 22. Water Math  
*c. 2.1 hrs  
d. 4 hrs  
37. A sludge pump is set to pump 5 minutes each hour. It pumps at the rate of 35 gpm. How many  
gallons of sludge are pumped each day?  
a. 120 gpd  
b. 175 gpd  
c. 840 gpd  
*d. 4200 gpd  
38. A primary clarifier has an influent suspended solids concentration of 250 mg/L. If the suspended  
solids removal efficiency is 60%, what is the primary effluent suspended solids concentration  
a. 10 mg/L  
*b. 100 mg/L  
c. 50 mg/L  
d. 150 mg/L  
39. A treatment plant receives a flow of 3.5 MGD. If the clarifier is 100 ft. long, 30 ft. wide, and 15 ft.  
deep, what is the surface loading rate?  
a. 78 gal/ft/day  
b. 700 gal/ft/day  
*c. 1,170 gal/ft/day  
d. 1,500 gal/ft/day  
e. 4,500 gal/ft /day  
40. A settling basin has a length of 60 ft, width of 20 ft, and depth of 12 ft At a flow rate of 60 MGD,  
what is the detention time of this basin?  
a. 15 minutes  
b. 30 minutes  
*c. 1.1 hours  
d. 2.3 hours  
41. If a clarifier has a capacity of 0.25 MG, what is the detention time in hours if it receives a flow of 3  
MGD  
a. 1.8 hrs  
*b. 2.0 hrs  
c. 2.25 hrs  
d. 2.50 hrs  
e. 3.00  
22.27 Dewatering math problems  
403  
Trickling Filter  
1. The influent to a trickling filter plant is 200 mg/L and the effluent BOD is 20 mg/L. What is the  
BOD removal efficiency (  
2. The flow to a trickling filter is 1.33 MGD. If the primary effluent has a BOD concentration of231  
mg/Land the trickling filter effluent has a BOD concentration of 83 mg/l, how many pounds of  
BOD are removed? (Ans: 1642 lbs/day)  
3. A 80-ft diameter trickling filter with a media depth of 7 ft receives a flow of 2,180,000 gpd. If the  
BOD concentration of the primary effluent is 139 mg/L, what is the organic loading on the trickling  
filter in lbs BOD/day/1000 cu ft? (Ans: 72 lbs/day-1000 ft3)  
4. A standard-rate filter, 90 ft in diameter, treats a primary effluent flow of 540,000 gpd. If the recir-  
culated flow to the trickling filter is 120,000 gpd, what is the hydraulic loading rate on the filter in  
gpd/sq ft? (Ans: 104)  
5. A 90 feet diameter trickling filter treats a 540,000 gpd primary effluent flow. If the recirculated flow  
to the trickling filter is 120,000 gpd, what is the hydraulic loading on the trickling filter in gpd/ft2.  
*a. 104 gpd/ft2  
b. 2 gpd/ft2  
c. 3 gpd/ft2  
d. 4 gpd/ft2  
6. A trickling filter plant operating at a recirculation ratio of 1 receives a raw wastewater flow of 2  
MGD. This means that the flow being applied to the filter would be:  
*a. 4mgd.  
b. 1mgd  
c. 3mgd  
d. 2mgd  
e. 6mgd  
7. What is the hydraulic loading of a trickling filter with a 100-foot diameter, if it receives a flow of  
4.0 MGD?  
a. 480 GPD/ft2  
*b. 510 GPD/ft2  
c. 540 GPD/ft2  
d. 570 GPD/ft2  
e. 600 GPD/ft2  
8. A 12 MGD primary clarifier effluent flow is fed to a trickling filter. If the total flow to the trickling  
filter including the recirculated flow is 17 MGD, what is the recirculation ratio?  
(Ans: 0.42)  
9. A trickling filter 100 ft. in diameter and 6 ft. deep receives a flow of 0.5 mgd with a BOD of 150  
mg/L. What is the BOD loading in pounds per day per 1000 cubic feet of filter media?  
a. 11.9.  
*b. 13.3.  
c. 26.8.  
d. 133.  
10. Calculate the organic load in lbs of BOD/day/1000 cu. ft. for a trickling filter 50 ft in diameter and  
eight (8) ft deep. Given a flow of 250,000 gpd, and a 150 mg/l BOD concentration of the primary  
effluent feed to the trickling filter.  
404  
Chapter 22. Water Math  
a. 2 lbs/day/1000 cu. ft.  
b. 5 lbs/day/1000 cu. ft.  
*c. 20 lbs/day/1000 cu. ft.  
d. 50 Ibs/day/1000 cu. ft.  
11. Calculate the pounds of BOD per day entering the trickling filter.  
DATA: Raw wastewater flow is 1.5 MGD.  
Raw wastewater BOD is 150 mg/L.  
There is a 30*a. 1600 lbs/day  
b. 880 lbs/day  
c. 870 lbs/day  
d. 560 lbs/day  
12. A trickling filter wastewater treatment plant receives a flow 1.95 MGD. Calculate the organic  
loading to this plant if it has a 135 ft diameter trickling filter with a 5 foot media depth and has a  
primary effluent BOD concentration of 110 mg/L.  
a. 0.5 lbs BOD/1,000 ft³/day.  
b. 2.7 lbs BOD/1,000 ft³/day.  
*c. 25 lbs BOD/1,000 ft³/day.  
d. 39 lbs BOD/1,000 ft³/day.  
e. 44 lbs BOD/1,000 ft³/day.  
13. At a 2.1 MGD wastewater treatment plant the influent suspended solids concentration to the pri-  
mary clarifier is 240 mg/l. The primary sludge contains 3.2gallons of primary sludge should be  
pumped per day?  
a. 63,000 gal/day ·  
b. 32,200 gal/day  
c. 8,200 gal/day  
*d. 7,550 gal/day  
14. Calculate the pounds of BOD per day entering the trickling filter  
DATA:  
Raw wastewater flow is 15 MGD  
Raw wastewater BOD is 150 mg/L  
There is a 30  
a. 560 lbs/day  
b. 870 1 bs/ day  
c. 880 1 bs /day  
*d. 1600 lbs/day  
15. Calculate the organic load in lbs of BOD/day/1000 cu ft for a trickling filter 50 ft in diameter and  
8 feet deep, a BOD design loading of 50 lbs/day/1000 cu ft, a flow of 250,000 gpd, and a primary  
effluent BOD to filter of 150 mg/L  
a. 2 lbs/day/1000 cu ft  
b. 5 lbs/day/1000 cu ft  
*c. 20 lbs/day/1000 cu ft  
d. 50 lbs/day/1000 cu ft  
16. What is the trickling filter hydraulic loading if the influent flow is 3 MGD and the recirculation rate  
is 4:1?  
a. 3 mgd  
22.27 Dewatering math problems  
405  
b. 6mgd  
c. 9mgd  
d. 12mgd  
*e. 15 mgd  
f. None of the above  
17. Calculate the organic load in lbs of BOD/day/1000 cu. ft. for a trickling filter 50 ft in diameter and  
eight (8) ft deep. Given a flow of 250,000 gpd, and a 150 mg/l BOD concentration of the primary  
effluent feed to the trickling filter.  
a. 2 lbs/day/1000 cu. ft.  
b. 5 lbs/day/1000 cu. ft.  
*c. 20 lbs/day/1000 cu. ft.  
d. 50 Ibs/day/1000 cu. ft.  
18. The desired trickling filter recirculation ratio is 1.4. If the primary effluent flow is 4.4 MGD what is  
the trickling filter effluent flow that needs to be recirculated.  
(Ans: 6.2)  
Stabilization Ponds  
1. A 40 acre pond receives a 0.6 MGD flow. If the pond is operated at a depth of 4ft, what is the  
detention time (days) of this pond. (Ans: 87)  
2. The flow to a pond is 750 gpm. The pond width is 200 ft and length is 450 ft. If the BOD in the  
pond influent is 240mg/L, what is the organic loading to this pond in lbs-BOD/day/acre? (Ans:  
1046)  
3. Calculate the depth the liquid level must be lowered in the fall below the normal 3-foot depth to  
prepare for winter and provide 180 day storage. DATA: Wastewater stabilization ponds total area of  
13.8 acres (600,000 sq. ft.) Average winter wastewater flow to the ponds is 100,000 gpd. Maximum  
liquid storage depth is 5 feet. (Assume no loss of liquid by percolation or evaporation and that the  
ponds will fill to the maximum storage level of 5 feet.)  
a. 0.1 ft.  
b. 1.0 ft.  
*c. 2.0 ft.  
d. 4.0 ft.  
4. Compute the lagoon’s detention time (days) DATA: Surface area= 6.0 acres Average depth= 3.0  
feet Average evaporation exceeds precipitation by 0.4 in/day Average daily flow (influent)= 0.25  
MGD 1 acre (area)= 43,560 sq ft 1 cu ft contains 7.5 gal  
a. 24  
*b. 32  
c. 37  
d. None of the above  
5. A pond is 600 ft. long and 150 ft. wide. If the pond is 6 ft. deep, what is the volume in acre ft?  
(Ans: 13.9)  
6. A wastewater pond has a detention time of 30 days and is operated at a pond depth of 4 ft. What is  
the hydraulic loading on the pond?  
a. 7.5 in/day  
*b. 1.6 in/day  
406  
Chapter 22. Water Math  
c. 0.63 in/day  
d. 0.41 in.day  
e. 013 in/day  
7. The volume of an aerated pond is 700,000 cubic feet. Flow to the pond averages .050 MGD. How  
many days detention time does the pond provide?  
a. 10.5 days.  
b. 14 days.  
*c. 105 days.  
d. 140 days.  
8. Assume a community of 1,100 people has a lagoon. The flow averages 80,000 gallons per day. The  
flow is  
a. 14 gallons per capita.  
*b. 72 gallons per capita.  
c. 80 gallons per capita.  
d. 140 gallons per capita.  
9. What is the organic loading on a lagoon (lbs./day/acre) if the influent BOD averages 164 mg/l and  
the flow averages 0.67 MGD. The lagoon has two (2) identical square cells with each side of 800 ft.  
a. 29 lbs./ac.  
b. 23 lbs./ac.  
*c. 31 lbs./ac.  
d. 52 lbs./ac.  
10. A stabilization pond is 1100 ft. long, 600 ft wide, and is operated at a depth of 5 ft. It receives a  
flow of 500,000 gpd and which has an influent BOD of 185mg/L. Using this information do the  
following:  
(a) Convert the flow to the pond in units of acre-ft/day (Ans. 1.53 acre-ft/day)  
(b) Find the area of this pond in units of acres. (Ans. 540,000 ft3 and 13.9 acre-ft)(Ans. 15.2  
acres)  
(c) Find the volume of pond in units of acre-feet. (Ans. 75.8 acre-ft)  
(d) Calculate the pond detention time in days. (Ans. 44.9)  
(e) Calculate the hydraulic loading to the pond in units of inches per day. (Ans. 1.2 in/day)  
(f) Calculate the organic loading to the pond (lbs of BOD/day/acre) (Ans. 50.8 lbs BOD/day-  
acre)  
Solution:  
500,000gal  
ft3  
acreft  
acreft  
(a)  
= 1.53  
day  
7.48gal 43,560 ft3  
day  
acres  
(b) (1,100600) ft2 ∗  
(c) (1,1006005) ft3 ∗  
= 15.2acres  
43,560ft2  
acresft  
= 75.8acreft  
43,560ft3  
volume  
flow  
1
7.48gal  
(d) DT =  
= (1,1006005) ft3 ∗  
= 49.4days  
500,000fracgalday ft3  
.
"
#
in  
day  
Pond depth (in)  
512in  
= =  
(e) Pond hydraulic loading rate  
=
Volume  
Flow  
49.4  
Pond detention time  
in  
1.2  
day  
22.27 Dewatering math problems  
lbs BOD per day  
407  
(f) Organic loading =  
area (acres)  
0.5MGD 185mg/l 8.34 43,560ft2  
50.9lbs BOD  
dayacre  
=⇒  
=
1,100600ft2  
acre  
11. What is the surface area of a pond that is 4 feet deep, if it holds 30 million gallons?  
Solution:  
Pond Volume = Sur face AreaDepth =30MG = Sur face Area4ft  
acreft  
30MG3.069  
MG  
=Sur face Area (acres) =  
= 23 acre  
4ft  
12. The influent flow to a pond is 10,000 gallons/hour with a suspended solids concentration of 142mg/L  
in the raw wastewater. How many lbs of suspended solids are sent to the pond daily?  
Solution:  
lbsSS  
day  
gal 24hrs  
MG  
mg  
l
lbs SS  
8.34 = 284  
day  
= 10000  
142  
hr  
day 1000000gal  
13. A pond is 260 ft. long and 80 ft. wide. What is the area of this pond in acres?  
Solution:  
(26080) ft2 ∗  
acre  
= 0.48acre  
43,560 ft2  
14. A pond has a volume of 1,800,000 ft3. If the flow to the pond is 425 gpm, what is the pond deten-  
tion time in days?  
Solution:  
volume  
flow  
1
day  
1440min  
7.48gal  
DT =  
= 1,800,000ft3 ∗  
= 22days  
ft3  
gal  
425  
min  
15. Find pond hydraulic loading in inches/day when the depth of the pond is 6 ft. and the detention  
time is 30 days.  
Solution:  
flow  
Hydraulic Loading (HL) =  
area  
vol  
vol  
DT  
Detention time (DT) =  
=flow =  
flow  
Substituting for flow in the HL formula above:  
vol  
DT  
area  
vol  
areaDT  
pond depth  
DT  
vol  
area  
HL =  
or  
=HL =  
as  
= pond depth  
Pond depth (in)  
612 inches  
=
2.4in  
day  
Pond hydraulic loading rate =  
=
Volume  
Flow  
30 days  
Pond detention time  
16. The flow to a pond is 7.2MGD. If the pond diameter is 350 ft and the BOD in the pond influent is  
170mg/L, what is the organic loading to this pond in lbs BOD/day/acre?  
Solution:  
lbs BOD per day 7.2MGD 170mg/l 8.34 43,560ft2  
4,624lbs BOD  
dayacre  
Organic loading =  
=
=
area (acres)  
0.7853502 ft2  
acre  
17. The flow to a pond is 750,000 gpd. If the pond diameter is 100 ft and the BOD in the pond influent  
is 300 mg/L, what is the organic loading to this pond in lbs BOD/day/acre? (Ans. 10,412 lbs  
BOD/day-acre)  
Solution:  
lbs BOD per day  
0.75MGD 300mg/l 8.34 43,560ft2  
Organic loading =  
=
=
area (acres)  
0.7851002 ft2  
acre  
10,413lbs BOD  
dayacre  
408  
Chapter 22. Water Math  
18. A 40 acre wastewater treatment pond receives a flow of 0.6 MGD. If the pond is operated at a depth  
of 4ft. What is the detention time of this pond?  
Solution:  
Volume  
Flow  
(404)acreft  
gal  
ft3  
acreft  
day 7.48gal 43,560ft3  
Pond detention time =  
=
= 87 days  
0.6106  
19. A 2.5 acre stabilization pond is operated at a depth of five (5) ft. What is the pond detention time if  
the flow to the pond is 18,000 cu. ft/day?  
Solution:  
Volume  
Flow  
(2.55)acreft  
ft3 acreft  
Pond detention time =  
=
= 30 days  
18,000  
day 43,560ft3  
20. The influent to a pond is 1,200 gpm with a BOD concentration of 180mglL in the raw wastewater.  
How many lbs/day of BOD are sent to the pond daily? (Ans. 2,600 lbs BOD/day)  
21. The influent flow to a pond is 10,000 gallons/hour with a suspended solids concentration of 142mg/L  
in the raw wastewater. How many lbs of suspended solids are sent to the pond daily? (Ans. 284 lbs  
SS/day)  
22. A pond is 260 ft. long and 80 ft. wide. What is the area of this pond in ft2 and acres? (Ans. 20,800  
ft2)  
23. A pond has a volume of 1,800,000 ft3. If the flow to the pond is 425 gpm, what is the pond deten-  
tion time in days? (Ans. 22 days)  
24. Find pond detention time in days given the following:  
• Flow to pond = 250 gpm  
• Pond length = 125 ft .  
• Pond width = 40ft.  
• Pond depth = 7ft.  
(Ans. 0.7 days)  
25. Find hydraulic loading in inches/day for a pond given the following:  
• Pond depth = 12ft.  
• Pond volume = 1,400,000ft3  
• Pond flow = 1,000,000gal/day  
Solution:  
gal  
ft3  
day 7.48gal  
"
#
1,000,000  
in  
day  
Flow  
Area  
in  
ft  
in  
day  
Pond hydraulic loading rate  
=
=⇒  
12 = 13.8  
1,400,000ft3  
12 ft  
Note: The area of the pond was found by dividing the volume (1,000,000 ft3) by the pond depth (12ft)  
26. The flow to a pond is 7.2MGD. If the pond diameter is 350 ft and the BOD in the pond influent is  
170mg/L, what is the organic loading to this pond in lbs BOD/day/acre? (Ans. 0.3 lbs BOD/day-  
acre)  
27. The flow to a pond is 200,000 gph. The pond width is 650 ft and the length is 300 ft. If the BOD in  
the pond influent is 265mg/L, what is the organic loading to this pond in lbs BOD/day/acre? (Ans.  
2,370 lbs BOD/day-acre)  
28. The influent flow to a pond is 600 gpm with a BOD concentration of 150mg/L in the raw wastewa-  
ter. How many lbs of BOD are sent to the pond daily? (Ans. 1081 lbs BOD/day)  
29. The influent flow to a pond is 50,000 gallons/hour with a suspended solids concentration of 125mg/L  
in the raw wastewater. How many lbs of suspended solids are sent to the pond daily? (Ans. 1,251  
22.27 Dewatering math problems  
409  
lbs TSS/day)  
30. The flow to a pond is 60 gpm with a COD concentration of 240mg/L in the pond influent. How  
many lbs/day of COD are sent to the pond daily? (Ans. 173 lbs COD/day)  
31. A pond is 100 ft. long and 210 ft wide. What is the area of this pond in ft2 and acres?(Ans. 21,000  
ft2 and 0.48 acres)  
32. A pond has a diameter of 80ft. What is the area of this pond in ft2 and acres? (Ans. 5,024 ft2 and  
0.12 acres)  
33. A pond is 450 ft. long and 450 ft. wide. If the pond is 3 ft. deep, what is the volume in ft3 and acre  
ft? (Ans. 607,500 ft3 and 13.9 acre-ft)  
34. A pond has a diameter of 230 ft. and is 6 ft. deep. What is the volume in ft3 and acre ft of this  
pond? (Ans. 249,159 ft3 and 5.7 acre-ft)  
35. A pond has a diameter of 50 ft. and is 8 ft deep. What is the volume of this pond in ft3 and acre ft?  
(Ans. 15,700 ft3 and 0.36 acre-ft)  
36. A pond has a volume of 2,800,000 ft3. If the flow to the pond is 500 gpm, what is the detention  
time in days? (Ans. 29 days)  
37. A 75 acre facultative pond is operated at a depth of 5.5 feet. Flow to this pond averages 2.2 MGD.  
The influent pond BOD averages,195 mg/L. Calculate the following:  
(a) The detention time (Ans. 61 days)  
(b) The hydraulic loading (Ans. 1.08 in/day)  
(c) The organic loading (Ans. 47 lbs BOD/day-acre)  
38. How deep must a 60 acre facultative pond be operated in order to have a detention time of 45 days.  
Flow to the pond is 2.0 MGD. (Ans. 4.6 ft)  
39. Find pond detention time in days.  
• Flow to pond = 300,000 gpm  
• Pond width = 450 ft.  
• Pond length = 500 ft.  
• Pond depth = 5 ft.  
(Ans. 28 days)  
40. A 75 acre pond receives an influent flow of 5 MGD. If the operating level of the pond is to be raised  
from 4.0 ft to 5.5 ft, how long will it take for the pond level to rise to the new discharge level (Ans.  
7.3 days)  
41. Find the pond detention time in days if the flow to the pond is 600 gpd, and the pond is 450ft wide,  
500 ft long and operating at a depth of 5 ft.  
42. The flow to a pond is 500 gpm with a COD concentration of 240mg/L in the pond influent. How  
many lbs/day of COD are present to the pond daily?  
43. The flow to a pond is 864,000 gal/day. The suspended solids concentration in the pond influent is  
380mg/L. How many lbs/ day of suspended solids are sent to the pond daily?  
44. A 5 acre stabilization pond is operated at a depth of five (5) ft. What is the pond detention time if  
the flow to the pond is 30,000 cu. ft/day?  
Solution:  
Volume  
Flow  
(55)acreft  
ft3 acreft  
Pond detention time =  
=
= 36.3 days  
30,000  
day 43,560ft3  
45. What is the hydraulic loading rate (in/day) on a 20 acre pond if it receives a flow of 3.3 acre-ft/day.  
a. 1 in/day  
410  
Chapter 22. Water Math  
b. 12 in/day  
c. 0.165 in/day  
*d. 2 in/day  
46. It is found that at a wastewater treatment pond system 35 days are required to bring the pond sys-  
tem to its operations depth of 3.5 feet. Assuming that the pond system was empty when the filling  
of the pond was initiated and that a constant incoming flow rate fills the pond, what would be the  
hydraulic loading on this pond system ?  
a. 0.47 inches/day  
*b. 1.2 inches/day  
c. 2.1 inches/day  
d. 3.5 inches/day  
e. 10.0 inches/day  
Activated Sludge  
1. The aeration tank of a conventional activated sludge plant contains 9300 lbs of MLSS. Lab tests  
show that the MLSS are 79% volatile solids. The primary effluent contains 2900 lbs of BOD.  
Calculate the F to M ratio for this wastewater treatment plant. (Ans: 0.4)  
2. How many lbs of solids are in a 400,000 gallon aeration tank if the suspended solids concentration  
is 1200 mg/l? (Ans: 4003.0)  
3. Operational data is given below for a conventional activated sludge treatment plant:  
Influent flow: 2.5 mgd  
Influent BOD: 220 mg/L  
Influent TSS: 240 mg/L  
Primary BOD removal efficiency: 30%  
RAS SS: 8,500 mg/L  
RAS VSS: 6,630 mg/L  
RAS flow: 160%  
Aeration tank volume: 1.8 MG  
Secondary clarifier volume: 0.8 MG  
MLSS: 3,600 mg/L  
MLVSS: 2,800 mg/L  
WAS flow: 35,000 gpd  
AS Effluent TSS: 25 mg/L  
AS Effluent BOD: 19 mg/L  
Settleability results: 60 min= 300 ml/L  
Settleability results: 30 min= 320 ml/L  
1) Calculate the MCRT (8 points)  
2) Calculate the F/M Ratio (8 points)  
3) Calculate the SVI (8 points)  
4) Calculate the aeration basin detention time  
4. Calculate the F to M ratio given the following.  
Plant flow- 0.8 MGD  
Aerator vol- 200,000 gal.  
Clarifier Volume-175,000 gal  
22.27 Dewatering math problems  
411  
Primary eff BOD- 120 mg/L  
MLVSS conc. -1950 mg/L  
(Ans: 0.25)  
5. Quantify the amount (lbs) of MLVSS in the two (2) aeration tanks if each tank is 40 ft. long, 15  
ft. wide and operating at a water depth of 12 ft. The MLSS is 2500 mg/l and is found to be 69%  
volatile. (Ans: 1728 lbs)  
6. The aeration tank of a conventional activated sludge plant contains 12,000 lbs of MLSS. Lab tests  
show that the MLSS are 79% volatile solids. The primary effluent contains 3800 lbs of BOD.  
Calculate the F to M ratio for this wastewater treatment plant. (Ans: 0.4)  
7. An activated sludge plant with the following parameters treats 5 MGD flow.  
Aeration tank volume: 0.25 MG  
Clarifier volume: 0.15 MG  
MLSS: 3000 mg/l  
RAS concentration: 6500 mg/l  
Secondary effluent TSS: 16 mg/l  
MCRT: 6 days  
1. Calculate the lbs MLSS in the aeration system (Aeration tank + Clarifier)  
2. Calculate the lbs TSS/day in the final effluent - 5 points  
3. At the given MCRT, calculate the WAS solids removed from the system (lbs/day)  
4. Using the given value of WAS concentration, calculate the gallons per day of WAS  
8. In an aeration tank, the MLSS is 2500 mg/l and recorded 30-minute settling test for a 1-litre MLSS  
samples indicates 230 ml of settled solids. What is the sludge volume index? (Ans: 92)  
9. An activated sludge process treating a 3 MGD flow has a 1 MG volume aeration basin and a 0.25  
MG volume final clarifier. The MLSS is 2000 mg/l and contains 82% volatile solids. If the sec-  
ondary effluent suspended solids concentration is 20mg/l and the WAS concentration is 7500 mg/l  
and is wasted at a rate of 0.06 MGD. Calculate the MCRT. (Ans: 4.9)  
10. The desired F/M ratio at a particular activated sludge plant is 0.5 lbs BOD/lb MLVSS. If the 3  
MGD primary effluent flow has a BOD of 165 mg/l, how many lbs of MLVSS should be main-  
tained in the aeration tank. (Ans: 8,257)  
11. Operational data is given below for a conventional activated sludge treatment plant:  
Influent flow: 0.500 mgd  
RAS SS: 12600 mg/L  
RAS VSS: 9450 mg/L  
Influent BOD: 199 mg/L  
Influent TSS: 221 mg/L RAS flow: 160%  
Aeration tank volume: 350,000 gal WAS flow: 6433 gpd  
Secondary clarifier volume: 150,000 gal  
MLSS: 4200 mg/L Effluent TSS: 45 mg/L  
MLVSS: 3150 mg/L Effluent VSS: 36 mg/L  
Settleability results: 30 min= 360 ml/L Effluent BOD: 20 mg/L  
Settleability results: 60 min= 330 ml/L  
Calculate the MCRT  
Answer: 21 days  
Calculate the F/M Ratio  
Answer: 0.09  
412  
Chapter 22. Water Math  
Calculate the aeration tank detention time  
Answer: 6.5 hrs  
Calculate the SVI  
Answer: 88 mL/g  
12. Calculate the F to M ratio given the following.  
Plant flow- 0.8 MGD  
Aerator vol- 200,000 gal.  
Clarifier Volume-175,000 gal  
Primary eff BOD- 120 mg/L  
MLVSS conc. -1950 mg/L  
(Ans: 0.25)  
13. An operator ran a 30 minute settling test on a mixed liquor sample using a 1000 ml graduated  
cylinder. The mixed liquor settled to 22% of the cylinder volume. The lab determined that the  
MLSS concentration was 3380 mg/l. What was the Sludge Volume Index for the mixed liquor?  
Correct Answer: 65.0  
14. The desired F/M ratio is .35lbs BOD/day/lb MLVSS. If 2,100 lbs of BOD enter the aerator daily,  
how many lbs of MLVSS should be maintained in the aeration tank?  
a. 3430 lbs  
b. 420 lbs  
*c. 6000 lbs  
d. 735 lbs  
15. A sludge settleability test shows a reading 220 ml after 30 minutes of settling in a one liter gradu-  
ated cylinder. Lab testing on this mixed liquor shows a MLSS concentration of 1850 mg/L and a  
MLVSS concentration of 1440 mg/L. Calculate SVI for this mixed liquor sample.  
*a. 119  
b. 108  
c. 153  
d. 138  
16. Given the following data, calculate the mean cell residence time (MCRT) .  
DATA: Influent flow = 3.5 MGD. MLSS = 2850 mg/L. Waste sludge flow = 0.08 MGD. Total  
secondary system volume = 2.0 MG. Waste activated sludge suspended solids conc -6000mg/L.  
Final effluent suspended solids = 25 mg/L.  
a. 6 days.  
b. 8 days.  
*c. 10 days.  
d. 12 days.  
17. What is the Sludge Volume Index given the following: MLSS = 2800 mg/L; MLVSS 240 mg/l;  
Settled Volume after 30 minutes = 250 ml/L  
a. 0.09 ml/g  
*b. 89 ml/g  
c. 0.11 ml/g  
d. 113 ml/g  
18. Given the following data, calculate the suspended solids required to maintain a food to microorgan-  
ism (F/M) ratio of 0.3  
DATA:  
22.27 Dewatering math problems  
413  
Daily flow = 10 MGD.  
Average primary effluent BOD = 150mg/L  
Aeration tank capacity = 200,000 gal  
a. 2100 mg/L  
*b. 2500 mg/L  
c. 3300 mg/L  
d. 4200 mg/L  
19. Given the following data, calculate the mean cell residence time (MCRT)  
DATA:  
Influent flow= 35 MGD  
MLSS = 2850 mg/L  
Waste sludge flow = 0.08 MGD  
Total secondary system volume= 20 MG  
Waste activated sludge suspended solids conc. = 6000 mg/L  
Final effluent suspended solids = 25 mg/L  
a. 6 days  
b. 8 days  
*c. 10 days  
d. 12 days  
20. Calculate the waste sludge pumping rate if you want to waste 2,000 pounds per day with a concen-  
tration of return sludge at 5000 mg/L  
a. 29 gpm  
*b. 33 gpm  
c. 35 gpm  
d. 37 gpm  
21. Calculate the return activated sludge (RAS) concentration  
DATA:  
SVI = 140  
SSV(30) = 350  
Influent flow = 25 MGD  
MLSS concentration = 2500 mg/L  
a. 8,300 mg /L  
*b. 7,100 mg /L  
c. 4,150 mg/L  
d. 2,800 mg/L  
22. Calculate the return activated sludge (RAS) concentration.  
DATA:  
SVI = 140 SSV = 350 30  
Influent flow= 2.5 MGD  
MLSS concentration= 2500 mg/L.  
a. 8300 mg /l.  
*b. 7100 mg/l.  
c. 4150 mg/l.  
d. 2800 mg/l.  
23. Given the following data, calculate the mean cell residence time (MCRT):  
414  
Chapter 22. Water Math  
DATA: Influent flow= 3.5 MGD.  
MLSS = 2850 mg/L.  
Waste sludge flow= 0.08 MGD.  
Total secondary system volume= 2.0 MG.  
Waste activated sludge suspended solids conc.: 6000 mg/L.  
Final effluent suspended solids= 25 mg/L.  
a. 6 days.  
*b. 8 days.  
c. 10 days.  
d. 12 days.  
24. Calculate the return activated sludge (RAS) concentration.  
DATA:  
SVI = 140 SSV = 350 30  
Influent flow= 2.5 MGD  
MLSS concentration= 2500 mg/L.  
a. 8300 mg /l.  
*b. 7100 mg/l.  
c. 4150 mg/l.  
d. 2800 mg/l.  
25. Calculate the return activated sludge (RAS) concentration.  
DATA:  
SVI = 140 SSV = 350 30  
Influent flow= 2.5 MGD  
MLSS concentration= 2500 mg/L.  
a. 8300 mg /l.  
*b. 7100 mg/l.  
c. 4150 mg/l.  
d. 2800 mg/l.  
26. Given the following data, calculate the mean cell residence time (MCRT) .  
DATA: Influent flow = 3.5 MGD. MLSS = 2850 mg/L. Waste sludge flow = 0.08 MGD. Total  
secondary system volume = 2.0 MG. Waste activated sludge suspended solids conc -6000mg/L.  
Final effluent suspended solids = 25 mg/L.  
a. 6 days.  
b. 8 days.  
*c. 10 days.  
d. 12 days.  
27. At an activated sludge wastewater treatment plant receiving 3.25 MGD, the final effluent suspended  
solids concentration averages 21.2 mg/L. What would the calculated MCRT value be when the  
aeration basin carries 2,050 mg/L MLSS and wastes 0.0550 MGD. The waste activated sludge  
has a concentration of 7,980 mg/L. The aeration tank has a volume of 1.00 MG and the secondary  
clarifier has an operational volume of 0.250 MG.  
Solution:  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
22.27 Dewatering math problems  
415  
MLSS in aeration tank (lbs) = 120508.34 = 17,097lbs  
MLSS in clarifier (lbs) = 0.2520508.34 = 4,274.3lbs  
SS e f fluent (lbs/day) = 3.25MGD21.2mg/l 8.34 = 574.6lbs/day  
SS WAS (lbs/day) = 0.055MGD7,980mg/l 8.34 = 3,660.4lbs/day  
17,097.6+4,274.3  
Plugging in the values calculated above: MCRT(days) =  
574.6+3,660.4  
= 4.8 = 5 days  
28. Calculate the MCRT of an activated sludge plant given the following information.  
Plant flow- 4.25 MGD  
WAS conc-7980 mg/l  
Waste flow- 0.055 MGD  
RAS conc.- 7980 mg/l  
Aeration tank vol-1MG  
Clarifier vol- 0.25 MG  
Final eff TSS conc. - 21.2 mg/l  
MLSS conc.- 2050 mg/l  
Solution:  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
MLSS in aeration tank (lbs) = 120508.34 = 17097lbs  
MLSS in clarifier (lbs) = 0.2520508.34 = 4274.3lbs  
SS e f fluent (lbs/day) = 4.25MGD21.2mg/l 8.34 = 751.4lbs/day  
SS WAS (lbs/day) = 0.055MGD7980mg/l 8.34 = 3660.4lbs/day  
17097.6+4274.3  
Plugging in the values calculated above: MCRT(days) =  
= 4.8 = 5days  
751.4+3660.4  
416  
29. Calculate the MCRT given the following.  
Chapter 22. Water Math  
Plant flow - 1.8 MGD  
MLSS conc - 2800 mg/l  
WAS flow - 0.04 MGD  
MLVSS conc. - 2190 mg/l  
Aerator vol - 0.3 MG  
Reactor vol. - 0.2 MG  
RAS conc. - 8150 mg/l  
Effluent SS conc.-18 mg/l  
Solution:  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
MLSS in aeration tank (lbs) = 0.328008.34 = 7005.6lbs  
MLSS in clarifier (lbs) = 0.228008.34 = 4670.4lbs  
SS e f fluent (lbs/day) = 1.8MGD18mg/l 8.34 = 270.2lbs/day  
SS WAS (lbs/day) = 0.04MGD8150mg/l 8.34 = 2718.8lbs/day  
7005.6+4670.4  
270.2+2718.8  
Plugging in the values calculated above: MCRT(days) =  
= 3.9 = 4days  
22.27 Dewatering math problems  
417  
30. In an aeration tank, the MLSS is 2650 mg/l and recorded 30-minute settling test indicates 221 ml/L.  
What is the sludge volume index?  
Solution:  
Settled sludge volume in ml/l after 30 min  
MLSS mg/l  
mg  
g
SVI (ml/g)=  
1000  
221ml/l  
2650mg/l  
mg  
g
SVI=  
1000  
= 83ml/g  
31. The desired F/M ratio is .35 lbs BOD/day/lb MLVSS. If 2,100 lbs of BOD enter the aerator daily,  
how many lbs of MLVSS should be maintained in the aeration tank?  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
2100  
x
=0.35 =  
=x = 6000lbs MLVSS  
32. A sludge settleability test shows a reading 220 ml after 30 minutes of settling in a one liter gradu-  
ated cylinder. Lab testing on this mixed liquor shows a MLSS concentration of 1850 mg/L and a  
MLVSS concentration of 1440 mg/L. Calculate SVI for this mixed liquor sample.  
*a. 119  
b. 108  
c. 153  
d. 138  
Solution:  
Settled sludge volume in ml/l after 30 min  
MLSS mg/l  
mg  
g
SVI (ml/g)=  
1000  
220ml/l  
1,850mg/l  
mg  
g
SVI=  
1000  
= 119ml/g  
33. Calculate F/M ratio based on the following data:  
Secondary influent BOD - 156 mg/l  
Four (4) aeration basins - 30 ft x 70 ft x 10 ft. deep  
Influent flow - 0.65 MGD  
MLSS - 3600 mg/l  
MLSS average % volatile - 72%  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
(lbs) MLVSS in the aeration tank  
F : M =  
F : M =  
1560.658.34  
= 0.06  
7.48gal  
MG  
36000.724(307010)ft3 ∗  
8.34  
1000000gal  
ft3  
418  
Chapter 22. Water Math  
34. An activated sludge plant operates well at an F:M ratio of between 0.23 and 0.28. Calculate the  
minimum MLSS concentration, given the following:  
Q = 0.4 MGD  
Primary influent BOD = 250 mg/l  
Primary effluent BOD = 128 mg/l  
Aeration tank vol. = 350,000 gallons  
Clarifier vol = 250,000 gallons  
MLSS has 80% volatile solids  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
1
=F : M ∝  
=F:M is inversely proportional to MLSS  
MLSS concentration  
So to have minimum MLSS conc. F:M needs to be the maximum of the range provided  
0.41288.34  
If the MLSS concentration = x: =F : M = 0.28 =  
=x = 5446mg/l MLSS  
0.35x0.8  
35. Given that an activated sludge plant with an influent flow of 1.2 MGD is operated at an MCRT of 6  
days and the parameters below, calculate the WAS flow rate (wasting rate) in gallon per day.  
Two aeration tanks – 0.5 MG each  
Two final clarifiers – 0.25 MG each  
WAS = 7500 ppm  
mg  
l
Final effluent = 20  
mg  
MLSS volatile solids content = 80%  
MLSS =3600  
L
36. Your plant currently is running at an MCRT of 5 days. You want to adjust things so that the plant  
is operating at an MCRT of 9 days. What is the new volume in gallons you should waste everyday  
to achieve this? Given the following information: Aeration tank volume of 750,000 gal. MLSS of  
2400 mg/l. Plant effluent SS of 7 mg/l, and a WAS concentration of 6500 mg/l. Influent flow 1.0  
MGD. [Ans. 29,292 gallons]  
37. Your engineer is upgrading your pond by providing surface aerators. She suggests you need two  
7-1/2 horsepower surface aerators with an oxygen transfer rate of 1.6 pounds of oxygen per horse-  
power per hour. Assume only 80% of the oxygen transfer will reduce the BOD 1:1. What is the  
minimum number of hours these time-clocked aerators will run to remove 85% of the influent  
BOD, which averages 295 pounds a day? [Ans. 13.1 hrs]  
38. Given the following information calculate MCRT for your WWTP. The plant is an extended aera-  
tion system with an aeration tank that has a volume of 66800 cu. ft. The MLSS is 1950 mg/l. The  
flow to the WWTP is 1.5 MGD and your effluent TSS is 9 mg/l. The WAS is 7000 mg/l and you  
waste 11990 gallons each day. [Ans. 10 days]  
22.27 Dewatering math problems  
419  
39. Given a system that has 25,000 pounds of MLSS in it and the desired quantity is 23,500 pounds of  
MLSS, how many gallons of sludge must be wasted if the RAS and WAS concentration is 7,200  
mg/l?[Ans. 24,980 gallons]  
40. Your aeration tank is 80 feet long, 40 feet wide and has 10.5 feet of mixed liquor depth. The  
MLVSS in the aeration tank is 1,850 mg/l. You are experiencing biological bulking and have read  
2 pounds of chlorine per 1,000 pounds of MLVSS should control this form of bulking. How many  
pounds of chlorine are added to the RAS portion of the RAS/WAS splitter box to achieve this  
process control strategy?[Ans. 7.8 lbs Cl2]  
41. An extended aeration plant is as follows: 0.28 MGD of 210 mg/l influent BOD, aeration basin  
volume 0.32 MG. The MLVSS is 4,450 mg/l (81% volatile) and the RAS is 9,200 mg/l. (a) What is  
the F/M? [Ans. 0.051 F:M] (b) How many pounds of suspended solids must be wasted to obtain an  
F/M of .06? [1785.6 lbs MLSS] (c) What is the new MLVSS in mg/l? [Ans. 3781 mg/l MLSS]  
42. Plant data: WAS Rate – 9 gpm Flow – 1.5 MGD MLSS – 2,350 mg/l Influent BOD – 210 mg/l  
RAS/WAS – 5,875 mg/l Influent TSS – 186 mg/l 30 min. Settling test – 320 ml/L Primary effluent  
BOD – 143 mg/l MLSS is 85% volatile solution Primary effluent TSS – 86 mg/l Effluent BOD – 18  
mg/l Aeration basins (2) Total Volume 468,750 gal Effluent TSS – 13 mg/l a). Calculate the F/M  
ratio: [Ans. 0.23] b). Calculate the sludge age: [Ans. 8.54days] c). Calculate MCRT: [Ans. 11.5  
days] d). Calculate the SVI: [Ans. 136]  
43. The 1.5 MGD influent to a treatment plant has an average BOD of 280 mg/L. 32% of this BOD  
is removed in the primary sedimentation tank. The primary effluent flows into an aeration tank  
containing 7000 lbs of MLSS with 79 % volatile matter. Calculate this plant’s F to M ratio.  
mg  
l
mg  
l
Influent BOD concentration to the AS basin: 280  
(10.32) = 190.4  
mg  
l
1.5MGD190.4  
8.34  
F
F
=
= 0.43  
M
70000.79  
M
44. Given that an activated sludge plant with an influent flow of 1.2 MGD is operated at an MCRT of 6  
days and the parameters below, calculate the WAS flow rate (wasting rate) in gallon per day.  
Two aeration tanks – 0.5 MG each  
Two final clarifiers – 0.25 MG each  
WAS – 7500 ppm  
mg  
Final effluent = 20  
l
mg  
L
MLSS volatile solids content = 80%  
MLSS –3600  
lbsMLSS(system)  
MCRT=  
lbs  
day  
lbs  
day  
E f fluentSS+  
WASSS  
mg  
L
lbs MLSS (system)= (20.5+20.25)MG3600  
8.34 = 45036lbs  
lbs  
day  
mg  
L
E f fluentSS = 1.2MG20  
8.34 = 200.2lbs  
45036  
lbs  
MCRT: 6days =  
lbs  
day day  
200.2  
+
WASSS  
lbs  
lbs  
day  
45036  
6
WASSS =  
200.2 = 7306  
day  
lbs  
7306  
75008.34  
7306  
= WASFlow(MGD)75008.34 =WASFlow(MGD) =  
= 0.116MGD =  
day  
420  
Chapter 22. Water Math  
gal  
day  
116,000  
45. Calculate the MCRT of an activated sludge plant given the following information.  
Plant flow- 4.25 MGD  
WAS conc-7980 mg/l  
Waste flow- 0.055 MGD  
RAS conc.- 7980 mg/l  
Aeration tank vol-1MG  
Clarifier vol- 0.25 MG  
Final eff TSS conc. - 21.2 mg/l  
MLSS conc.- 2050 mg/l  
Solution:  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
MLSS in aeration tank (lbs) = 120508.34 = 17097lbs  
MLSS in clarifier (lbs) = 0.2520508.34 = 4274.3lbs  
SS e f fluent (lbs/day) = 4.25MGD21.2mg/l 8.34 = 751.4lbs/day  
SS WAS (lbs/day) = 0.055MGD7980mg/l 8.34 = 3660.4lbs/day  
17097.6+4274.3  
Plugging in the values calculated above: MCRT(days) =  
= 4.8 = 5days  
751.4+3660.4  
22.27 Dewatering math problems  
421  
46. Calculate the MCRT given the following.  
Plant flow - 1.8 MGD  
MLSS conc - 2800 mg/l  
WAS flow - 0.04 MGD  
MLVSS conc. - 2190 mg/l  
Aerator vol - 0.3 MG  
Reactor vol. - 0.2 MG  
RAS conc. - 8150 mg/l  
Effluent SS conc.-18 mg/l  
Solution:  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
MLSS in aeration tank (lbs) = 0.328008.34 = 7005.6lbs  
MLSS in clarifier (lbs) = 0.228008.34 = 4670.4lbs  
SS e f fluent (lbs/day) = 1.8MGD18mg/l 8.34 = 270.2lbs/day  
SS WAS (lbs/day) = 0.04MGD8150mg/l 8.34 = 2718.8lbs/day  
7005.6+4670.4  
270.2+2718.8  
Plugging in the values calculated above: MCRT(days) =  
= 3.9 = 4days  
422  
Chapter 22. Water Math  
47. In an aeration tank, the MLSS is 2650 mg/l and recorded 30-minute settling test indicates 221 ml/L.  
What is the sludge volume index?  
Solution:  
Settled sludge volume in ml/l after 30 min  
MLSS mg/l  
mg  
g
SVI (ml/g)=  
1000  
221ml/l  
2650mg/l  
mg  
g
SVI=  
1000  
= 83ml/g  
48. The desired F/M ratio is .35 lbs BOD/day/lb MLVSS. If 2,100 lbs of BOD enter the aerator daily,  
how many lbs of MLVSS should be maintained in the aeration tank?  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
2100  
x
=0.35 =  
=x = 130,100lbs MLVSS  
49. Calculate F/M ratio based on the following data:  
Secondary influent BOD - 156 mg/l  
Four (4) aeration basins - 30 ft x 70 ft x 10 ft. deep  
Influent flow - 0.65 MGD  
MLSS - 3600 mg/l  
MLSS average % volatile - 72%  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
1560.658.34  
F : M =  
= 0.06  
7.48gal  
MG  
36000.724(307010)ft3 ∗  
8.34  
1000000gal  
ft3  
22.27 Dewatering math problems  
423  
50. An activated sludge plant operates well at an F:M ratio of between 0.23 and 0.28. Calculate the  
minimum MLSS concentration, given the following:  
Q = 0.4 MGD  
Primary influent BOD = 250 mg/l  
Primary effluent BOD = 128 mg/l  
Aeration tank vol. = 350,000 gallons  
Clarifier vol = 250,000 gallons  
MLSS has 80% volatile solids  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
1
=F : M ∝  
=F:M is inversely proportional to MLSS  
MLSS concentration  
So to have minimum MLSS conc. F:M needs to be the maximum of the range provided  
0.41288.34  
If the MLSS concentration = x: =F : M = 0.28 =  
=x = 5446mg/l MLSS  
0.35x0.8  
51. 5. Given that an activated sludge plant with an influent flow of 1 MGD is operated at an MCRT of  
6 days and the parameters below, calculate the WAS flow rate (wasting rate) in gallon per day. (10  
points)  
Two aeration tanks – 0.45 MG each  
Two final clarifiers – 0.2 MG each  
WAS = 6500 ppm  
mg  
l
Final effluent = 16  
mg  
MLSS volatile solids content = 85%  
MLSS =3000  
L
Solution:  
lbsMLSS(system)  
MCRT=  
lbs  
day  
lbs  
E f fluentSS+  
WASSS  
day  
mg  
L
lbs MLSS (system)= (20.45+20.2)MG3000  
8.34 = 32,526lbs  
lbs  
day  
mg  
L
E f fluentSS = 1MG16  
8.34 = 133.4lbs  
32,526  
lbs lbs  
day day  
MCRT: 6days =  
133.4  
+
WASSS  
lbs  
day  
32,526  
lbs  
WASSS =  
133.4 = 5,288  
day  
6
lbs  
day  
5,288  
= WASFlow(MGD)65008.34  
424  
Chapter 22. Water Math  
5,288  
6,5008.34  
gal  
day  
=WASFlow(MGD) =  
= 0.097546MGD = 97,546  
52. Operational data is given below for a conventional activated sludge treatment plant:  
Influent flow: 2.5 mgd  
Influent BOD: 220 mg/L  
Influent TSS: 240 mg/L  
Primary BOD removal efficiency: 30%  
Aeration tank volume: 1.8 MG  
Secondary clarifier volume: 0.8 MG  
MLSS: 3,600 mg/L  
MLVSS: 2,800 mg/L  
RAS SS: 8,500 mg/L  
RAS VSS: 6,630 mg/L  
RAS flow: 100%  
WAS flow: 35,000 gpd  
AS Effluent TSS: 25 mg/L  
AS Effluent BOD: 19 mg/L  
Settleability results: 60 min= 300 ml/L  
Settleability results: 30 min= 320 ml/L  
(a) Calculate the MCRT  
(b) Calculate the F/M Ratio  
(c) Calculate the SVI  
Solution:  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
(a) MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
MLSS in aeration tank (lbs) = 1.83,6008.34 = 54,043lbs  
MLSS in clarifier (lbs) = 0.83,6008.34 = 24,019lbs  
SS e f fluent (lbs/day) = 2.5MGD25mg/l 8.34 = 521lbs/day  
35,000  
SS WAS (lbs/day) =  
MGD8,500mg/l 8.34 = 2,481lbs/day  
1,000,000  
54,043+24,019  
521+2,481  
MCRT(days) =  
= 26 days  
(lbs/day) primary e f fluent BOD entering the aeration tank  
(lbs) MLVSS in the aeration tank  
(b) F : M =  
F : M =  
220(10.3)2.58.34  
= 0.08  
1.82,8008.34  
Settled sludge volume in ml/l after 30 min  
MLSS mg/l  
mg  
g
(c) SVI (ml/g)=  
1000  
320ml/l  
3,600mg/l  
mg  
g
SVI=  
1000  
= 89ml/g  
53. What is the Sludge Volume Index given the following:  
MLSS = 2800 mg/L; MLVSS 2400 mg/l; Settled Volume after 30 minutes = 250 ml/L  
22.27 Dewatering math problems  
425  
Solution:  
Settled sludge volume in ml/l after 30 min  
mg  
g
SVI (ml/g)=  
1000  
MLSS mg/l  
250ml/l  
2800mg/l  
54. Plant Influent Flow: 3.5 MGD  
mg  
g
SVI=  
1000  
= 89ml/g  
Plant Influent BOD: 276 mg/l  
Primary Treatment BOD Removal: 36%  
Desired F/M: 0.3  
Find lbs of MLVSS that should be maintained in the activated sludge treatment process. (Answer:  
17,187 lbs)  
55. Given that an activated sludge plant with an influent flow of 1 MGD is operated at an MCRT of 6  
days and the parameters below, calculate the WAS flow rate (wasting rate) in gallon per day  
Two aeration tanks – 0.45 MG each  
Two final clarifiers – 0.2 MG each  
WAS = 6500 ppm  
mg  
Final effluent = 16  
l
mg  
MLSS volatile solids content = 85%  
MLSS =3000  
L
(Answer: 97,546 gal/day)  
56. The aeration tank of a conventional activated sludge plant contains 9300 Ibs of MLSS. Lab tests  
show that the MLSS are 79% volatile solids. The primary effluent contains 2900 Ibs of BOD.  
Calculate the F to M ratio for this wastewater treatment plant. (Answer: 0.4)  
57. How many lbs of solids are in a 400,000 gallon aeration tank if the suspended solids concentration  
is 1200 mg/l? (Answer: 4003 lbs)  
58. An activated sludge wastewater treatment plant treats an average flow of 3.25 MGD. The aeration  
tank has a volume of 1.0 MG and a MLSS concentration of 2100 mg/L. The final clarifier has an  
operating volume of 0.25 MG. Effluent suspended solids concentration averages 21.2 mg/L. The  
WAS concentration is 8100 mg/L and is wasted at the rate of 0.055 MGD. Calculate the MCRT.  
59. What is the F/M ratio on an aeration tank if 1,500 pounds of BOD are added per day and 5000  
pounds of volatile solids are under aeration?  
60. In an aeration tank, the MLSS is 2500 mg/l and recorded 30-minute settling test indicates 230 ml.  
What is the sludge volume index?  
61. An activated sludge aeration tank receives a primary effluent flow of 2.61 MGD with a BOD con-  
centration of 195 mg/L. The mixed liquor volatile suspended solids concentration is 2560 mg/L and  
the aeration tank volume is 470,000 gallons. What is the current F/M ratio?  
62. Calculate the MCRT given the following. Plant flow- 2.8 MGD  
MLSS conc.- 2800 mg/L  
WAS flow- 0.07 MGD  
MLVSS conc.- 2190mg/L  
Aerator volume - 0.38MG  
RAS SS concentration - 6800 mg/L  
Clarifier volume - 0.19 MG  
Final effluent TSS - 18 mg/L  
63. If in a conventional activated sludge treatment plant the aeration tank contains 7000 lbs of MLSS  
and the final clarifier contains 2500 lbs of MLSS. If 1300 lbs of solids are wasted each day and 120  
426  
Chapter 22. Water Math  
lbs of solids leave in the final effluent, Calculate the MCRT.  
64. The aeration tank of a conventional activated sludge plant contains 9300 Ibs of MLSS. Lab tests  
show that the MLSS are 79% volatile solids. The primary effluent contains twenty-nine hundred  
pounds (2900 Ibs) of BOD. Calculate the F to M ratio for this wastewater treatment plant.  
65. A conventional activated sludge plant receives an average flow of 2.5 MGD. The influent BOD  
averages 230 and the primary effluent BOD average 160 mg/I. The 0.6 MG aeration tank has a  
MLSS cone. of 2800 mg/L and a MLVSS cone. of 21 50 mg/L. Calculated the F to M ratio for this  
plant.  
66. A convention activated sludge plant receives an average influent flow of 800,000 gpd. The influent  
BOD averages 200 mg/L and on the average 32% of this BOD is removed in the primary sedimen-  
tation tank. The mixed liquor tank contains 3250 Ibs of MLSS. These solids contain 77% volatile  
matter. Calculate this plant’s F to M ratio.  
67. In an conventional activated sludge plant calculations show that the aeration tank contains 6500 Ibs  
of MLSS and the final clarifier contains 2500 Ibs of MLSS. Twelve hundred and fifty pounds (1250  
Ibs) of solids is wasted each day and 100 Ibs/day of solids leave in the final effluent. Calculate the  
MCRT for this plant.  
68. A activated sludge plant has a total of 25,000 pounds of solids in the system (i.e. aerator + final  
clarifier) and has an average effluent flow of 3.75 MGD. The effluent suspended solids concentra-  
tion averages 22 mg/L. The WAS has a concentration of 7500 mg/L and is being wasted at 0.05  
MGD. Calculate the MCRT.  
69. An activated sludge wastewater treatment plant discharges an average flow of 3.25· MGD. The  
mixed liquor tank has a volume of 1.0 MG and a MLSS concentration of 2100 mg/L. The final clar-  
ifier has an operating volume of 0.25 MG. Effluent suspended solids concentration averages 21.2  
mg/L. The WAS concentration is 8100 j mg/L and is wasted at the rate of 0.055 MGD. Calculate  
the MCRT.  
70. In an conventional activated sludge plant show that the aeration tank contains 6500 Ibs of MLSS  
and the final clarifier contains 2500 lbs of MLSS. 1250 Ibs of solids is wasted each day and 100  
Ibs/day of solids leave in the final effluent. Calculate the MCRT for this plant. Answer: 7 days  
71. An activated sludge plant has a total of 25,000 pounds of solids in the system (aerator + final Clar-  
ifier) and has an average effluent flow of 3.75 MGD. The effluent suspended solids concentration  
averages 22 mg/L. The WAS has a concentration ’of 7500 .mg/L and is being wasted at 0.05 MGD.  
Calculate the MCRT. Answer: 7 days  
72. An activated sludge wastewater treatment plant discharges an average flow of 3.25 MGD. The  
mixed liquor tank has a volume of 1.0 MG and a MLSS concentration of 2100 mg/L. The final clar-  
ifier has an operating volume of 0.25 MG. Effluent suspended solids concentration averages 21.2  
mg/L and the WAS concentration is 8100 mg/L and is wasted at the rate of 0.055 MGD. Calculate  
the MCRT. Answer: 5.1 days  
73. A conventional activated sludge plant receives an average flow of 2.5 MGD. The influent BOD  
averages 230 and the primary effluent BOD averages 160mg/L The 0.6 MG aeration tank has a  
MLSS conc. of 2800 mg/L and a MLVSS conc. of 2150mg/L. Calculate the F to M ratio for this  
plant. Answer: 0.31  
74. A conventional activated sludge plant receives an influent flow of 800,000 gpd. The influent BOD  
averages 200 mg/L and on the average 32% of this BOD is removed in the primary sedimentation  
tank. The mixed liquor tank contains 3250 Ibs of MLSS. These solids contain 77% volatile matter.  
Calculate this plant’s F to M ratio. Answer: 0.36  
22.27 Dewatering math problems  
427  
75. Calculate the MCRT given the following. Plant flow- 1.8 MGD  
MLSS conc.- 2800 mg/L  
WAS flow- 0.04 MGD  
MLVSS conc.- 2190mg/L  
Aerator vol- 0.3MG  
RAS conc.- 8150 mg/L  
Clarifier vol- 0.lMG  
Final eff TSS- 26 mg/L  
Answer: 3 days  
76. Calculate the F to M ratio given the following Information. Plant flow- 3.7 MGD  
Primary eff TSS- 120mg/L  
Aerator Vol - 0.85 MG  
MLSS conc.- 3200 mg/L  
Clarifier Vol - 0.25 MG  
MLVSS conc.- 2550 mg/L  
Primary effluent BOD-120 mg/L  
Answer: 0.2  
77. Calculate the F to M ratio given the following. Plant flow- 0.8 MGD Aerator vol- 200,000 gal.  
Clarifier Volume-175,000 gal Primary eff BOD- 120 mg/L MLVSS cone. -1950 mg/L  
Answer: 0.25  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
1200.88.34  
F : M =  
= 0.25  
19500.28.34  
78. Calculate the MCRT of an activated sludge plant given the following information. Plant flow- 4.25  
MGD  
WAS conc-7980 mg/L  
Waste flow- 0.055 MGD  
RAS conc.- 7980 mg/L  
Aeration tank vol-1MG  
Clarifier vol- 0.25 MG  
Final eff TSS conc. - 21.2 mg/L  
MLSS conc.- 2050 mg/L  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
MLSS in aeration tank (lbs) = 120508.34 = 17097lbs  
MLSS in clarifier (lbs) = 0.2520508.34 = 4274.3lbs  
SS e f fluent (lbs/day) = 4.25MGD21.2mg/l 8.34 = 751.4lbs/day  
SS WAS (lbs/day) = 0.055MGD7980mg/l 8.34 = 3660.4lbs/day  
17097.6+4274.3  
Plugging in the values calculated above: MCRT(days) =  
= 4.8 = 5days  
751.4+3660.4  
79. Calculate the MCRT given the following.  
Plant flow - 1.8 MGD  
428  
Chapter 22. Water Math  
MLSS conc - 2800 mg/L  
WAS flow - 0.04 MGD  
MLVSS conc. - 2190 mg/L  
Aerator vol - 0.3 MG  
Reactor vol. - 0.2 MG  
RAS conc. - 8150 mg/L  
Effluent SS conc.-18 mg/L  
Solution:  
MLSS in aeration tank (lbs)+MLSS in clari fier (lbs)  
MCRT(days) =  
SS e f fluent (lbs/day)+SS WAS (lbs/day)  
MLSS in aeration tank (lbs) = 0.328008.34 = 7005.6lbs  
MLSS in clarifier (lbs) = 0.228008.34 = 4670.4lbs  
SS e f fluent (lbs/day) = 1.8MGD18mg/l 8.34 = 270.2lbs/day  
SS WAS (lbs/day) = 0.04MGD8150mg/l 8.34 = 2718.8lbs/day  
7005.6+4670.4  
270.2+2718.8  
Plugging in the values calculated above: MCRT(days) =  
= 3.9 = 4days  
80. The desired F/M ratio is .35 lbs BOD/day/lb MLVSS. If 2,100 lbs of BOD enter the aerator daily,  
how many lbs of MLVSS should be maintained in the aeration tank?  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
2100  
x
=0.35 =  
=x = 6000lbs MLVSS  
81. Calculate F/M ratio based on the following data:  
Secondary influent BOD - 156 mg/l  
Four (4) aeration basins - 30 ft x 70 ft x 10 ft. deep  
Influent flow - 0.65 MGD  
MLSS - 3600 mg/l  
MLSS average % volatile - 72%  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
1560.658.34  
F : M =  
= 0.06  
7.48gal  
MG  
36000.724(307010)ft3 ∗  
8.34  
1000000gal  
ft3  
82. An activated sludge plant operates well at an F:M ratio of between 0.23 and 0.28. Calculate the  
minimum MLSS concentration, given the following: Q = 0.4 MGD Primary influent BOD = 250  
mg/L Primary effluent BOD = 128 mg/L Aeration tank vol. = 350,000 gallons Clarifier vol =  
250,000 gallons MLSS has 80% volatile solids Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
1
=F : M ∝  
MLSS concentration  
22.27 Dewatering math problems  
429  
So to have minimum MLSS conc. F:M needs to be the maximum of the range provided  
0.41288.34  
If the MLSS concentration = x: =F : M = 0.28 =  
=x = 5446mg/l MLSS  
0.35x0.8  
83. If the target MCRT for an activated sludge plant with the following parameters is 8 days, what is  
the required wasting rate (in gallons per day)? (8 points)  
Aeration Tank Volume = 1.5 MG  
Clarifier Volume = 1.2 MG  
MLSS Concentration = 2500 mg/l  
MLSS Volatile Solids = 79%  
WAS SS = 5100 mg/l  
Secondary Effluent SS = 12 mg/l  
Plant flow = 15 MGD  
(Total MLSS lbs in the aeration system(aeration tank + clari fier))  
MCRT =  
lbs  
day  
(Total amount in  
of SS leaving the system(e f fluent +WAS))  
(MLSSaeration tank(lbs) + MLSSclarifier (lbs))  
(SSeffluent(lbs/day)+SSWAS(lbs/day))  
MCRT =  
mg  
l
(1.5+1.2)MG8.342500  
=8days =  
mg  
l
mg  
l
(15MGD12  
8.34)(lbs/day SSeffluent +(xMGD WAS5100  
8.34)(lbs/day SSWAS  
)
(1.5+1.2)MG8.342500  
(15128.34(lbs/day)SSeffluent +xMGD WAS51008.34)(lbs/day)SSWAS  
(1.5+1.2)MG8.342500  
=
8
xMGD WAS51008.34(lbs/day)SSWAS  
5535.7  
=
(15128.34(lbs/day)SSeffluent  
8
gal  
42,534x = 5535.7 =x =  
= 0.1301MGD = 130,100  
42,534  
day  
84. Given the following: Plant Influent Flow: 3.5 MGD  
Plant Influent BOD: 276 mg/l  
Primary Treatment BOD Removal: 36%  
Desired F/M: 0.3  
Find lbs of MLVSS that should be maintained in the activated sludge treatment process. (7 points) F :  
lbs In fluent BOD  
lbs MLVSS  
276(10.36)3.58.34  
M =  
=0.3 =  
=lbs MLVSS = 17,187lbs  
lbsMLVSS  
5. Given that an activated sludge plant with an influent flow of 1 MGD is operated at an MCRT of  
6 days and the parameters below, calculate the WAS flow rate (wasting rate) in gallon per day. (10  
points)  
Two aeration tanks – 0.45 MG each  
Two final clarifiers – 0.2 MG each  
WAS = 6500 ppm  
mg  
l
Final effluent = 16  
mg  
MLSS volatile solids content = 85%  
MLSS =3000  
L
Solution:  
lbsMLSS(system)  
MCRT=  
lbs  
day  
lbs  
E f fluentSS+  
WASSS  
day  
mg  
L
lbs MLSS (system)= (20.45+20.2)MG3000  
8.34 = 32,526lbs  
430  
Chapter 22. Water Math  
lbs  
day  
mg  
L
E f fluentSS = 1MG16  
8.34 = 133.4lbs  
32,526  
lbs lbs  
day day  
MCRT: 6days =  
133.4  
+
WASSS  
lbs  
day  
32,526  
lbs  
WASSS =  
133.4 = 5,288  
day  
6
lbs  
day  
5,288  
= WASFlow(MGD)65008.34  
5,288  
gal  
day  
=WASFlow(MGD) =  
= 0.097546MGD = 97,546  
6,5008.34  
85. The desired F/M ratio is 0.35. If 2,100 lbs of BOD enter the aerator daily and the volume of the  
mixed liquor in the aeration tank is 375,000 gallons , what should be the target MLVSS concentra-  
tion?  
Solution:  
(lbs/day) primary e f fluent BOD entering the aeration tank  
F : M =  
(lbs) MLVSS in the aeration tank  
2100  
x
=0.35 =  
=x = 6000lbs MLVSS  
=6,000lbs MLVSS = 8.341,000,000 MGxmg  
375,000  
MLVSS  
L
mg MLVSS  
=xmg  
=
= 1,900  
MLVSS  
L
6000  
8.340.375  
L
Solids Treatment Math Problems  
1. Calculate the volatile solids reduction in an anaerobic digester given the following information:  
Raw sludge feed to digester: 73.7 % VS and digested sludge: 57.2 %VS  
*a. 52.3%  
b. 64%  
c. 16.5%  
d. 42%  
Solution:  
The VS reduction of the digester is provided by the Van Kleeck equation  
VSin VSout  
Digester VS reduction(%) =  
100  
VSin VSin VSout  
0.7370.572  
Digester VS reduction(%) =  
100 = 52.3%  
0.7370.7370.572  
2. 42,000 gallons of 6% sludge containing 67% volatile matter is pumped to the digester. The di-  
22.27 Dewatering math problems  
431  
gester reduces the volatile matter by 52%. What volume of sludge in gallons containing 5% solids  
remains after digestion? [Ans: 32,841 gal]  
3. An anaerobic digester is 37’ in diameter and 27’ deep with a 5,000 gallon daily sludge flow. The  
sludge is 6% solids and 66% volatile solids. What is the volatile solids loading in pounds per cubic  
foot per day? [0.057 lbs VS/ft3-day]  
4. An Imhoff cone result is 5.5 ml/l. Approximately how many gallons of primary sludge will need to  
be pumped if the flow is 0.7 MGD? [3,850 gallons]  
5. A sludge digester equipped with a floating cover is 31 ft. inside diameter. The corbels are 16 feet  
above the floor and the top of the wall is 20 feet above the floor. The digester currently has a sludge  
depth of 16.5 ft. How many gallons of sludge are needed to displace the cover 1 foot? [5,643  
galllons]  
6. 10,000 gallons/day of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). If  
50% of the VS is destroyed, how many lbs of VS is destroyed per day?  
Solution:  
1,168lbs VS destroyed  
10,000 Gal 8.34 lbs sludge 0.040.7 lbs VS feed 0.5 lbs VS destroyed  
=
day  
Gal  
lb sludge  
lbs VS feed  
day  
7. Two sludges are blended together as follows: 15,000 gal. primary sludge at 4.1% solids. 28,000 gal.  
secondary sludge at 1.3% solids. What is the combined solids concentration? [Ans. 2.28%] If the  
primary sludge is 68% VS and the secondary sludge is 63% VS, how many pounds of VS are in the  
combined sludge? [Ans. 5400.3 lbs VS]  
8. 12,000 gallons of 1.8% sludge is pumped to a thickener, and is thickened to 3800 gallons of 4.9%  
solids. The supernatant is returned back to the head of the STP for treatment. What is the volume in  
gallons and solids concentration in ppm of the supernatant?[Ans. 24,980 gallons]  
9. How many pounds of solids are pumped to a digester each day if the digester receives 10,000 gpd  
of sludge at 5% solids concentration?  
Solution:  
lbs TS 10,000gal sludge (8.340.05lbsTS)  
= 4,170  
lbs TS  
day  
=
day  
day  
gal sludge  
10. Calculate the % VS reduction in a digester given the volatile solids content of the influent sludge to  
the digester is 70% and the volatile solids content of the sludge leaving the digester is 52.5%  
0.70.525  
Solution: Digester VS reduction(%) =  
100 = 53%  
0.70.70.525  
11. If an anaerobic digester receives sludge with VS of 80% and discharges digested sludge with a 60%  
VS. Its VS reduction is:  
a. 20%  
b. 25%  
c. 63%  
d. 89%  
Solution:  
0.80.6  
Digester VS reduction(%) =  
100 = 63%  
0.80.80.6  
12. Calculate the VS loading to the digester in lbs/day if 10,000 gallons of sludge containing 5% TS  
with and average VS content of 78%  
Solution:  
Digester VS loading (lbs/day)  
432  
Chapter 22. Water Math  
10,000 gallons sludge 8.34lbs sludge 0.050.78lbsVS  
=
= 3,253lbs sludge per day  
day  
gal  
lb sludge  
13. Primary sludge containing five percent (5%) solids is pumped to a digester continuously at a rate  
of 25 gpm. How many pounds of volatile solids are added to the digester each day if the volatile  
solids content is 73% of the total solids?  
a. 1,310 lbs/day  
b. 1,800 lbs/day  
c. 9,830 lbs/day  
d. 10,960 lbs/day  
e. 15,010 lbs/day  
Solution:  
Digester VS loading (lbs/day)  
251,440 gallons sludge 8.34lbs sludge 0.050.73lbsVS  
=
= 10,960 lbs sludge per day  
day  
gal  
lb sludge  
14. An anaerobic digester is 37’ in diameter and 27’ deep with a 5,000 gallon daily sludge flow. The  
sludge is 6% solids and 66% volatile solids. What is the volatile solids loading in pounds per cubic  
foot per day?  
Solution:  
lbs VS  
Digester Loading  
day  
Digester volatile solids loading rate =  
Digester volume(V) ft3  
gal sludge  
day  
lbsVS  
gal sludge  
5000  
(8.340.060.66)  
lbs VS  
dayft3  
=
= 0.057  
π
(
372 27)ft3  
4
15. How much gas is produced in the above digester in ft3/day if the digested sludge contains 2.5%  
total solids of which 59% is volatile solids and the gas production rate is 14 ft3/lb VS destroyed?  
Solution:  
0.70.59  
Digester VS reduction(%) =  
100 = 38.3%  
0.700.700.59  
lbs VS reduction 3153lbs VS feed 0.383 VS reduction  
lbs VS reduction  
day  
=
= 1,208  
day  
day  
ft3gas produced  
day  
lbs VS reduced 14 ft3 gas produced  
ft3 digester gas produced  
= 1208  
= 16,912  
day  
lb VS reduced  
day  
16. The sludge feed to a digester is 80,000 gal/day. The sludge contains 4.5% total solids with 75%  
volatile solids. If 50% of VS are reduced in the digester:  
(a) Find lbs VS destroyed per 1000 gal of digester capacity per day if the digester radius is 55 ft  
with an operating sludge level of 25 ft.(5 points)  
22.27 Dewatering math problems  
433  
Solution:  
gal  
ft3  
Digester Volume: (π 552 25) ft3 7.48  
= 1,776,220gallons = 1,776.2 (1000 gallons)  
lbs  
gal  
lbsVS  
lb  
lbs VS reduction  
0.5  
lb VS  
80,000gal 8.34  
(0.0450.75)  
lbs VS reduction  
day  
=
day  
lbs VS reduction  
day  
= 11,259  
lbs VS reduction  
day  
1,776.2 (1000 gallons)  
11,259  
lbs VS reduction  
1000 digester capacity  
lbs VS reduction  
1000gallons digester volume  
=
= 6.3  
(b) What is the digester gas production in Btu/day? Assume 14 cu. ft digester gas per lb of VS  
destroyed and a 650 Btu/cu. ft heating value for the digester gas produced (5 points)  
Solution:  
ft3gas produced  
day  
lbs VS reduced 14 ft3 gas produced  
ft3 digester gas produced  
day  
= 11,259  
= 157,626  
day  
lb VS reduced  
BTU produced  
ft3 gas produced 650BTU gas produced  
BTU produced  
= 157,626  
= 102,456,900  
day  
day  
ft3gas  
day  
17. The volatile acids concentration of sludge in an anaerobic digester is 195 mg/l. If the maximum  
volatile acids:alkalinity ratio is 0.087, what should the alkalinity be in mg/l?  
Solution:  
195  
x
195  
0.087  
mg  
l
Volatile acid:Alkalinity=0.087=  
=x =  
= 2,241  
434  
Chapter 22. Water Math  
18. 10,000 gallons of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). 50% of  
the VS are destroyed, creating 15 ft3 of gas per lbs. of VS destroyed. How much gas is produced  
each day?  
lbs  
gal  
lbsVS 0.5lbs VS reduction  
10,000gal 8.34  
(0.040.7)  
lbs VS reduction  
day  
lbs VS reduction  
day  
lb  
lb VS  
=
= 1,168  
day  
ft3gas produced  
day  
lbs VS reduced 15 ft3 gas produced  
ft3 digester gas produced  
day  
= 1,168  
= 17,520  
day  
lb VS reduced  
19. Given a sludge flow of 5,500 gpd with 6.5% solids containing 72% VS to a digester with a VS  
destruction of 48% If the digester is digester is 40 ft. diameter with a sludge level of 25 ft, find  
lbs. VS destroyed per 1000 gal of digester capacity per day. What is the digester gas production in  
Btu/day if the % VS reduction in the digester is 58% and the digester raw sludge feed is 6000 cu.  
Ft/day containing 4.5% total solids with a 64% VS content? Assume 13.5 cu. ft digester gas per lb  
of VS destroyed and a 650 Btu/cu. ft heating value for the digester gas produced  
π
4
gal  
ft3  
Digester Volume: ( 402 25)ft3 7.48  
= 940,086gallons = 940 (1000 gallons)  
lbs  
gal  
lbsVS 0.48lbs VS reduction  
5,500gal 8.34  
(0.0650.72)  
lbs VS reduction  
day  
lb  
lb VS  
=
=
day  
lbs VS reduction  
day  
1,030  
lbs VS reduction  
day  
940 (1000 gallons)  
1,030  
lbs VS reduction  
1000 digester capacity  
lbs VS reduction  
1000gallons digester volume  
=
= 1.1  
ft3gas produced  
lbs VS reduced 13.5 ft3 gas produced  
ft3 digester gas produced  
day  
= 1,030  
= 13,905  
day  
day  
lb VS reduced  
BTU produced  
day  
ft3 gas produced 650BTU gas produced  
BTU produced  
= 13,905  
= 9,038,250  
day  
ft3gas  
day  
20. Two sludges are blended together as follows: 60,000 gal per day. primary sludge at 4.5% solids and  
28,000 gal secondary sludge at 5% solids.  
(a) What is the combined solids concentration? (3 points)  
60,0004.5+28,0005  
= 4.66%  
88,000  
(b) How many lbs of solids (per day) are in the combined sludge (3 points)  
88,000gal 8.340.0466 = 34,200lbs  
(c) If the primary sludge is 68% VS and the secondary sludge is 83% VS, how many pounds of  
VS are in the combined sludge? (3 points)  
22.27 Dewatering math problems  
8.340.0450.68lbs VS  
435  
8.340.050.83lbs VS  
60,000gal∗  
+28,000gal∗  
= 25,003lbs VS  
gal  
gal  
(d) What is the digester feed VS%? (3 points)  
25,003lbs VS  
34,200lbsTS  
= 73.1%  
(e) If the digester is 120’ diameter with a liquid height of 20’, what is the VS loading in lbs  
VS/ft3-day (3 points)  
25,003lbs VS  
0.7851202 20  
0.11lbs VS  
dayft3  
=
(f) If the digested sludge has 52% VS, calculate the digester VS reduction percent (3 points)  
0.731.52  
Digester VS reduction(%) =  
100 = 60.1%  
0.7310.7310.52  
(g) What is the gas production (ft3/day) if the digester produces 14 ft3 gas/lb VS destroyed (3  
points)  
lbs VSdestroyed 14 ft3 gas produced  
25,003*0.601  
= 210,375ft3gas  
lb VS destroyed  
(h) How many belt presses are needed to keep up with the digested sludge flow if the belt presses  
can be operated at a maximum flow of 50 GPM (3 points)  
gal sludge  
day  
day  
1440min  
88,000  
= 61GPM  
@50 GPM per press - 2 BFP required  
(i) If the digested sludge feed to the belt filter presses is 2.6% and assuming the belt press feed  
solids capture is 90%. How many lbs of solids (dry) are produced by the BFP (4 points)  
gal  
day  
lbs solids feed  
day  
lbs solids captured  
lbs solids feed  
88,000  
8.340.026  
0.9  
= 17,174lbs solids per day  
436  
Chapter 22. Water Math  
(j) If the BFP produces a 20% cake, how many wet tons cake produced per day (4 points)  
17,174lbs solids produced  
100lbs cake  
20lbs solids produced 2,000lbs  
tons  
42.9tons  
day  
=
day  
(k) If the cake density is 68 lbs/cu. ft, how much time will it take to fill a 30 cu. yd bin (3 points)  
(42.92000)lbs cake  
day  
59.583  
(Ans. 15.4 hours)  
=
min  
day  
1440min lbs cake  
59.583  
min∗  
ft3  
0.876 ft3  
=
lbs cake  
68lbs cake  
min  
27 ft3  
yd3  
min  
hrs  
30yd3 ∗  
0.876ft3 60min  
= 15.4hrs  
(l) What will be the cost of hauling dewatered cake per day @ $65 per ton cake (2 points)  
tons $65  
day ton  
$2,789  
day  
42.9  
=
21. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS  
with and average VS content of 76% is fed to the digester  
22. Calculate the organic loading rate to two 320,000 gallon anaerobic digesters given: Primary sludge  
feed rate of 25,000 gallons with 6.2% TS containing 73% solids and SG of 1.03. Secondary sludge  
feed rate of 30,500 gallons with 3.8% TS containing 77% solids (7 points) (Answer: 0.2lbs VS/day-  
ft3 )  
23. Calculate the organic loading rate to two 320,000 gallon anaerobic digesters given:  
Primary sludge feed rate of 25,000 gallons with 3.8% TS containing 73% solids and SG of 1.03.  
Secondary sludge feed rate of 30,500 gallons with 3.8% TS containing 77% solids  
lbs VS  
dayft3  
(Answer: 0.2  
)
24. 325 GPM of WAS with 7,000 mg/l TSS is thickened in a DAFT. Assuming an air:solids ratio of  
0.05 lb of air for each lb of feed solids, The SCFM (standard cubic feed per minute) air required to  
thicken the WAS given 1 SCFM = 0.075 lb is:  
(Answer: 12.6 SCFM)  
Solution:  
Correct Answer:  
0.075 lbs air  
x SCFM  
lb air  
lb solids  
SCF  
0.05 =  
=
325  
1,000,000  
(
MG per min70008.34) lbs solids  
0.0532570008.34  
0.0751,000,000  
=x SCFM =  
= 12.6 SCFM Correct answer  
Incorrect Answer #1:  
12,600SCFM  
22.27 Dewatering math problems  
437  
Incorrect Answer #2:  
0.126SCFM  
Incorrect Answer #3:  
1.26SCFM  
25. 30,000 gpd of 4.5% sludge is dewatered in a centrifuge yielding 24.5 yd3/day of 25% cake with a  
density of 65 lb per ft3. Calculate the percentage solids recovery.  
(Answer: 95.5%)  
BTU  
day  
26. Estimate the amount of heat value of the gas produced from an anaerobic digester in  
the following:  
given  
Raw sludge pumping schedule  
Sludge pumping rate  
Raw sludge %TS  
12 min/hr  
68 GPM  
5.2%  
Raw sludge %VS  
Gas production  
72.5%  
12 ft3  
lb VS destroyed day  
Percent CO2  
34%  
1%  
Other gases  
BTU  
Pure methane net heat value  
932  
ft3  
BTU  
day  
(Answer: 23,146,220  
)
27. 12,000 ft3 of anaerobically digested sludge containing 2.8% TS is dewatered in a centrifuge. The  
centrifuge yields 37 yd3 of 26% of dewatered cake with a density of 73 lb/ft3. Calculate the solids  
capture rate.  
Solution:  
gal (8.340.028lbs TS)  
= 20,960lbs TS  
lbs TS feed to centrifuge = 12,000ft3 sludge7.48  
ft3  
gal sludge  
ft3 (73lbs0.26 TS)  
lbs TS feed from centrifuge = 37yd3 sludge27  
= 18,961lbs TS  
yd3  
ft3 sludge  
18,961lbs solids produced by centri fuge  
solids capture rate =  
100 = 90.4%solids capture  
20,960lbs solids fed from digester  
28. At a 60 GPM of 2.8% feed a belt press which has a 90% solids capture rate produces a 20% cake at  
68 lbs/ ft3. How long would it take to fill a 3 yd3 bin  
Solution:  
Approach:  
• First calculate the amount of cake solids produced in terms of weight per time.  
• From the weight of the cake produced calculate the volume time, and finally  
using the calculated value of the volume of the cake produced per time, calculate the time  
required to fill the bin  
cake TS produced lbs 60gallons sludge 8.34lbs sludge feed 0.028lbs TS  
=
0.9  
lb sludge  
min  
min  
galllon  
12.61lbs TS  
=
min  
ft3 cake produced  
min  
12.61lbs TS 100lbs cake  
ft3 cake  
68lbs cake  
0.927 ft3cake  
=
=
min  
20lbs TS  
min  
438  
Chapter 22. Water Math  
ft3 cake produced  
min  
12.61lbs TS 100lbs cake  
ft3 cake  
68lbs cake  
0.927 ft3cake  
=
=
min  
20lbs TS  
min  
min  
0.927 ft3  
27 ft3  
yd3  
Time required to fill the bin =  
3yd3 ∗  
= 75min  
22.27 Dewatering math problems  
439  
29. Calculate the annual cost savings for an improvement of cake solids from 20% to 21% if 400 wet  
tons (@ 20% cake) are produced each day and the average cake hauling and disposal cost is $65  
per wet ton.  
400 wet tons 20tons solids  
Total (dry) solids produced per day =  
= 80tons solids  
day  
100wet tons  
80 tons solids 100wet tons  
Tons wet cake @21% solids =  
day  
21tons solids  
380tons wet solids  
=
day  
wet tons  
day  
days  
yr  
$65  
wet ton  
Net savings ($/yr) = (400380)  
365∗  
= $474,500 per year  
NOTE: General formula for future reference: So if the cake dryness goes up from 20% to  
26% and currently a utility is spending $1,000,000 per year for biosolids hauling and disposal,  
their net savings will be: (26-20)/26*1,000,000 = $230,769. Conversely say if the dewatering  
cake solids drops to 18%, the net impact will be: (18-20)/18*1,000,000 = $-111,111 (loss or  
extra cost).  
30. Two sludges are blended together as follows: 60,000 gal per day. primary sludge at 4.5% solids and  
28,000 gal secondary sludge at 5% solids.  
(a) What is the combined solids concentration? (3 points)  
60,0004.5+28,0005  
= 4.66%  
88,000  
(b) How many lbs of solids (per day) are in the combined sludge (3 points)  
88,000gal 8.340.0466 = 34,200lbs  
(c) If the primary sludge is 68% VS and the secondary sludge is 83% VS, how many pounds of  
VS are in the combined sludge? (3 points)  
8.340.0450.68lbs VS  
8.340.050.83lbs VS  
60,000gal∗  
+28,000gal∗  
= 25,003lbs VS  
gal  
gal  
(d) What is the digester feed VS%? (3 points)  
25,003lbs VS  
34,200lbsTS  
= 73.1%  
(e) If the digester is 120’ diameter with a liquid height of 20’, what is the VS loading in lbs  
VS/ft3-day (3 points)  
25,003lbs VS  
0.7851202 20  
0.11lbs VS  
dayft3  
=
(f) If the digested sludge has 52% VS, calculate the digester VS reduction percent (3 points)  
0.731.52  
Digester VS reduction(%) =  
100 = 60.1%  
0.7310.7310.52  
(g) What is the gas production (ft3/day) if the digester produces 14 ft3 gas/lb VS destroyed (3  
points)  
440  
Chapter 22. Water Math  
= 210,375ft3gas  
lbs VSdestroyed 14 ft3 gas produced  
25,003*0.601  
lb VS destroyed  
(h) How many belt presses are needed to keep up with the digested sludge flow if the belt presses  
can be operated at a maximum flow of 50 GPM (3 points)  
gal sludge  
day  
day  
1440min  
88,000  
= 61GPM  
@50 GPM per press - 2 BFP required  
(i) If the digested sludge feed to the belt filter presses is 2.6% and assuming the belt press feed  
solids capture is 90%. How many lbs of solids (dry) are produced by the BFP (4 points)  
gal  
day  
lbs solids feed  
day  
lbs solids captured  
lbs solids feed  
88,000  
8.340.026  
0.9  
= 17,174lbs solids per day  
(j) If the BFP produces a 20% cake, how many wet tons cake produced per day (4 points)  
17,174lbs solids produced  
100lbs cake  
20lbs solids produced 2,000lbs  
tons  
42.9tons  
day  
=
day  
(k) If the cake density is 68 lbs/cu. ft, how much time will it take to fill a 30 cu. yd bin (3 points)  
(42.92000)lbs cake  
day  
59.583  
(Ans. 15.4 hours)  
=
min  
day  
1440min lbs cake  
59.583  
min∗  
ft3  
0.876 ft3  
=
lbs cake  
68lbs cake  
min  
27 ft3  
yd3  
min  
hrs  
30yd3 ∗  
0.876ft3 60min  
= 15.4hrs  
(l) What will be the cost of hauling dewatered cake per day @ $65 per ton cake (2 points)  
tons $65  
day ton  
$2,789  
day  
42.9  
=
31. 12,000 ft3 of anaerobically digested sludge containing 2.8% TS is dewatered in a centrifuge. The  
centrifuge yields 37 yd3 of 26% of dewatered cake with a density of 73 lb/ft3. Calculate the solids  
capture rate.  
Solution:  
gal (8.340.028lbs TS)  
= 20,976lbs TS  
lbs TS feed to centrifuge = 12,000ft3 sludge7.48  
ft3  
gal sludge  
ft3 (73lbs0.26 TS)  
lbs TS feed from centrifuge = 37yd3 sludge27  
= 18,961lbs TS  
yd3  
ft3 sludge  
18,961lbs solids produced by centri fuge  
solids capture rate =  
100 = 90.4%solids capture  
20,976lbs solids fed from digester  
32. At a 60 GPM of 2.8% feed a belt press which has a 90% solids capture rate produces a 20% cake at  
68 lbs/ ft3. How long would it take to fill a 3 yd3 bin  
Solution:  
Approach:  
• First calculate the amount of cake solids produced in terms of weight per time.  
• From the weight of the cake produced calculate the volume time, and finally  
using the calculated value of the volume of the cake produced per time, calculate the time  
required to fill the bin  
cake TS produced lbs 60gallons sludge 8.34lbs sludge feed 0.028lbs TS  
=
0.9  
lb sludge  
min  
min  
galllon  
22.27 Dewatering math problems  
441  
12.61lbs TS  
=
min  
ft3 cake produced  
min  
12.61lbs TS 100lbs cake  
ft3 cake  
68lbs cake  
0.927 ft3cake  
=
=
=
=
min  
20lbs TS  
min  
ft3 cake produced  
min  
12.61lbs TS 100lbs cake  
ft3 cake  
68lbs cake  
0.927 ft3cake  
min  
20lbs TS  
min  
min  
0.927 ft3  
27 ft3  
yd3  
Time required to fill the bin =  
3yd3 ∗  
= 75min  
442  
Chapter 22. Water Math  
33. Calculate the annual cost savings for an improvement of cake solids from 20% to 21% if 400 wet  
tons (@ 20% cake) are produced each day and the average cake hauling and disposal cost is $65  
per wet ton.  
Solution:  
400 wet tons 20 tons solids  
Total (dry) solids produced per day =  
= 80 tons solids  
day  
100 wet tons  
80 tons solids 100 wet tons  
Tons wet cake @21% solids =  
day  
21 tons solids  
380 tons wet solids  
=
day  
wet tons  
day  
days  
yr  
$65  
wet ton  
Net savings ($/yr) = (400380)  
365∗  
= $474,500 per year  
NOTE: General formula for future reference: So if the cake dryness goes up from 20% to  
26% and currently a utility is spending $1,000,000 per year for biosolids hauling and disposal,  
(2620)  
their net savings will be:  
$1,000,000 = $300,000  
20  
34. A belt press is used for dewatering 70 GPM digested sludge containing 3% TS, seven hours per day.  
At the end of the day it produces 24 yd3 of 16% TS biosolids @ 65 lbs/ft3 density. What is the  
percent belt press solids recovery?  
Solution:  
70 gal 8.34 lbs 0.03 lbs solids 760 min 7,356 lbs solids  
lbs solids feed to belt press:  
=
min  
gal  
lb feed sludge  
day  
day  
24 yd3 cake 27 ft3 65 lbs cake 0.16 lbs solids 6,739 lbs solids  
lbs solids in belt press cake:  
=
day  
yd3  
ft3  
lb cake  
day  
6,739  
7,356  
belt press solids recovery:  
= 0.92 or 92%  
35. Calculate the air required (SCFM) to meet a 0.04 lb air:lb feed solids ratio for a 100 GPM WAS  
flow with a solids content of 6500mg/l? Assume 0.08 lbs air/SCF air.  
Solution:  
0.08 lbs air  
x SCF per minute  
lb air  
SCF  
0.04 =  
=
100  
lb solids  
(
MG per min6500mg/l 8.34) lbs solids  
1,000,000  
0.04  
0.0148  
=0.04 = 0.0148 x SCF per minute =x SCF per minute =  
= 2.7 SCFM  
36. A treatment plant receives an influent flow of 30 MGD with a TSS concentration of 280 mg/l. The  
primary treatment removes 55% TS and the primary sludge is pumped to a 40 ft diameter gravity  
thickener. Calculate the average solids loading to the thickener in lbs TSS/day-ft2  
Solution:  
(30 MGD 2800.55 mg/l 8.34)lbs TSS per day  
Solids loading to gravity thickener=  
=
0.785402 ft2  
30.7 lbsTSS/dayft2  
22.27 Dewatering math problems  
443  
37. Two, 110’ diameter digesters with a cone depth of 15 ft and operating at a cylindrical liquid level of  
28 ft is fed with a blend of primary and secondary sludge. The 85,000 gal/day primary sludge feed  
contains 5% solids with a 65% VS content. The secondary sludge flow is 33,000 gallons per day  
with a 5.5% solids and 86% VS content.  
(a) What is the combined solids concentration?  
Combined solids concentration:  
C1 V1 +C2 V2 = C3 V3  
C1 V1 +C2 V2 C1 V1 +C2 V2 585,000+5.533,000  
= 5.14% TS content  
=C3 =  
=
=
V3  
V1 +V2  
88,000+33,000  
(b) What is the digester VS loading rate (lbs VS/ ft3)  
lbs VS  
Digester VS loading =  
Digester volume  
8.34(Pri. Sldg gal/dayPri.Sldg TS%VS+ Sec. sludge gal/daySec. sludgeTS%VS)  
=⇒  
=⇒  
h
i
3.14D2 h1  
2∗  
+0.785D2 h  
12  
8.34(88,000 gal/day0.050.65 + 33,000 gal/day0.0550.86)  
lbs VS  
ft3  
= 0.059  
h
i
(3.141102 15  
2∗  
+0.7851102 28  
12  
(c) If the digested sludge has 52% VS, calculate the digester VS reduction percent  
lbs VS  
lbs TS  
Calculate VS% in: =  
100  
8.34(Pri. Sldg gal/dayPri.Sldg TS%VS+ Sec. sludge gal/daySec. sludgeTS%VS)  
8.34(Pri. Sldg gal/dayPri.Sldg TS+ Sec. sludge gal/daySec. sludgeTS)  
=⇒  
=⇒  
8.34(88,000 gal/day0.050.65 + 33,000 gal/day0.0550.86)  
100  
8.34(88,000 gal/day0.05 + 33,000 gal/day0.055)  
36,870lbs VS  
=
51,833 lbs TS 100 = 71.1%VSin  
0.71.52  
Digester VS reduction(%) =  
100 = 55.8% VS reduction  
0.710.710.52  
(d) What is the gas production (ft3/day) if the digester produces 14 ft3 gas/lb VS destroyed  
14 ft3 gas produced  
lb VS destroyed  
36,870 lbs VSin 0.558lbs VS destroyed∗  
= 288,028 ft3 gas produced  
(e)  
(f) How many belt presses are needed to keep up with the digested sludge flow if the belt presses  
can be operated at a maximum flow of 50 GPM  
gal sludge  
day  
day  
1440min  
121,000  
= 84GPM  
@50 GPM per press - 2 BFP required  
(g) If the digested sludge feed to the belt filter presses is 2.6% and assuming the belt press feed  
solids capture is 90%. How many lbs of solids (dry) are produced by the BFP  
gal  
day  
lbs solids feed  
day  
lbs solids captured  
lbs solids feed  
121,000  
8.340.026  
0.9  
= 23,028lbs solids per day  
(h) If the belt press produces a 20% cake at 68 lbs/ ft3. How long would it take to fill a 3 yd3 bin?  
444  
Chapter 22. Water Math  
lbs cake  
lbs solids  
day  
lb cake  
0.2 lb solids  
lbs cake : 23,028  
= 115,140  
day  
lbs cake  
ft3  
yd3  
day  
yd3 cake  
hr  
Volume of cake produced: 115,140  
= 2.61  
day  
68 lbs cake 27 ft3 24 hrs  
= 1.15 hrs  
hr  
(i) Time to fill the 3 cu. yd bin: 3 yd3 ∗  
2.61yd3 cake  
eak  
38. Two sludges are blended together as follows: 60,000 gal. primary sludge at 4.5% solids and 28,000  
gal. secondary sludge at 5% solids.  
(a) What is the combined solids concentration?  
(b) If the primary sludge is 68% VS and the secondary sludge is 83% VS, how many pounds of  
VS are in the combined sludge?  
(c) If the digester is 120’ diameter cylinder with a liquid height of 20’, what is the VS loading in  
lbs VS/ft3-day?  
(d) If the digester VS reduction is 60%, what is the gas production (ft3/day) if the digester  
produces 14 ft3 gas/lb VS destroyed?  
(e) How many belt presses are needed to keep up with the digested sludge flow if the belt presses  
are operated at a maximum flow of 50 GPM ?  
(f) Knowing the digester VS reduction and the TS feed, calculate the solids (TS) concentration  
coming out of the digesters  
(g) Assuming the belt press solids capture is 90%; if it produces a 20% cake which is 68 lbs/ft3  
in how many hours will it take to fill a 3 yd3 bin with the cake?  
Solution:  
,
(a) What is the combined solids concentration?  
Combined solids concentration: C1 V1 +C2 V2 = C3 V3  
C1 V1 +C2 V2 C1 V1 +C2 V2  
4.560,000+528,000  
= 4.66%  
=C3 =  
=
=
V3  
V1 +V2  
60,000+28,000  
60,000  
mg  
lbs TS from primary sludge: (4.510,000) TS∗  
MGD8.34 = 22,518 lbs TS  
MGD8.34 = 11,676 lbs TS  
l
1,000,000  
28,000  
1,000,000  
mg  
lbs TS from secondary sludge: (510,000) TS∗  
l
Total TS = 22,518+11,676=34,194 lbs TS  
34,194 lbs TS  
mg  
l
TS conc.(mg/l) =  
= 46,591  
or 4.66%  
88,000  
1,000,000  
8.34∗  
MGD  
(b) If the primary sludge is 68% VS and the secondary sludge is 83% VS, how many pounds of  
VS are in the combined sludge?  
lbs of VS in combined sludge: CVS1 V1 +CVS2 V2 = CVS3 V3  
CVS1 V1 +CVS2 V2 CVS1 V1 +CVS1 V2  
=CVS3  
=CVS3  
=
=
=
V3  
V1 +V2  
4.560,0000.68+528,0000.83  
= 3.406%  
60,000+28,000  
88,000  
lbs VS = 3.4110,000mg/l ∗  
8.34 = 25,027 lbs VS  
1,000,000  
mg  
60,000  
1,000,000  
lbs VS from primary sludge: (4.50.6810,000) VS∗  
MGD8.34 = 15,312 lbs VS  
l
mg  
28,000  
lbs VS from secondary sludge: (5 0.83 10,000) VS ∗  
MGD 8.34 =  
l
1,000,000  
9,691 lbs VS Total VS = 15,312+9,691= 25,003 lbs VS  
(c) If the digester is 120’ diameter cylinder with a liquid height of 20’, what is the VS loading in  
lbs VS/ft3-day?  
22.27 Dewatering math problems  
445  
lbs VS per day  
Digester volume ( ft3)  
(d) If the digester VS reduction is 60%, what is the gas production (ft3/day) if the digester  
produces 14 ft3 gas/lb VS destroyed?  
25,003 lbs VS per day  
0.7851202 20 ft3  
0.11 lbs VS  
dayft3  
VS loading =  
=
=
Digester gas production = lbs VS reducedgas production rate(ft3/lb lbs VS reduced  
14 ft3 gas  
lb VS destroyed  
=25,0030.6 lbs VSreduced ∗  
= 210,025 ft3 gas per day  
(e) How many belt presses are needed to keep up with the digested sludge flow if the belt presses  
are operated at a maximum flow of 50 GPM?  
Flow to the belt press=Flow to the digesters as digesters are fixed volume tanks  
=flow in to the digester = flow out of the digesters - which will be the flow to the belt press  
gal  
day  
day 1440 min  
=@50 GPM belt press flow capacity need 2 belt presses  
gal  
min  
=flow to the belt press = 88,000 gal =88,000  
= 61  
446  
Chapter 22. Water Math  
(f) Knowing the digester VS reduction and the TS feed, calculate the solids (TS) concentration  
coming out of the digesters  
Approach: Calculate the lbs of VS destroyed in the digester and subtract that out of the lbs  
TS feed to the digester. This will be the mass of solids leaving the digester. The mass of  
solids divided by the volume of the feed (also the out) would be the TS concentration  
34,194 lbs TS feed 25,0030.6 lbs VS reduced  
=⇒  
lbs  
106lbs  
88,000 gal 8.34  
gal 1,000,000lbs  
lbs solids  
= 20,441  
or 26,150 ppm or 2.61%  
106 lbs  
(g) Assuming the belt press solids capture is 90%; if it produces a 20% cake which is 68 lbs/ft3  
in how many hours will it take to fill a 3 yd3 bin with the cake?  
Approach  
,
i. Calculate the lbs of solids feed to the belt press.  
ii. Based on the 90% capture rate, calculate the lbs of solids leaving the belt press.  
iii. Knowing the cake is 20% solids, calculate the lbs of cake produced.  
iv. Knowing the density of the cake (68 lbs/ ft3), calculate the volume of the cake produced.  
v. Knowing the cake volume produced, calculate the time it will take to fill the 3 cu. yd bin  
Calculations:  
i. (34,194 lbs TS feed 25,0030.6 lbs VS reduced) solids feed belt press  
= 19,192 lbs TS solids feed  
ii. solids leaving belt press: 19,192*0.9=17,273 lbs  
lbs solids  
day  
lb cake  
lbs cake  
day  
iii. lbs cake : 17,273  
= 86,365  
0.2 lb solids  
lbs cake  
ft3  
yd3  
day  
iv. Volume of cake produced: 86,365  
=
day  
68 lbs cake 27 ft3 24 hrs  
yd3 cake  
1.96  
hr  
hr  
v. Time to fill the 3 cu. yd bin: 3 yd3 1.96  
= 1.53 hrs  
yd3 cake  
Question 1: A sludge thickened from 1% to 4% solids will be reduced in volume by how much?  
Solution: Assume 1 lbs of solids are in the sludge.  
100 lbs sludge  
Sludge weight for 1% solids = 100 lbs because, 1% =⇒  
1 lb solids  
100 lbs sludge  
4 lb solids  
25 lbs sludge  
1 lb solids  
Sludge weight for 4% solids = 25 lbs because, 4% =⇒  
Thus, sludge weight and volume of the 4% sludge will be 25% of the original 1% sludge  
or  
Alternatively, C1(1%) V1(1%) = C2(4%) V2(4%) =0.01V1(1%) = 0.04V2(4%) =V2(4%)  
0.25V1(1%)  
=
Thus, volume of the 4% sludge will be 25% of the original 1% sludge.  
39. If a belt filter press is used 14 hrs per day to dewater 90 GPM digester sludge feed with 2.7% solids,  
the daily lbs solids feed is:  
Solution:  
lbs solids feed to belt press:  
gal  
min  
lbs  
gal  
lbs solids  
lb feed sludge  
min  
day  
lbs solids  
day  
90  
8.34  
0.027  
(1460)  
= 17,024  
22.27 Dewatering math problems  
447  
40. A belt filter press is used 10 hrs per day to dewater 80 GPM digester sludge feed with 2.5% solids.  
The belt press produces 22% biosolids (bulk density - 70 lbs/ft3) which is loaded into trucks for  
offsite disposal. Five truckloads of biosolids are produced every three days with each truckload  
averaging 13.3 yd3 of material.  
(a) Calculate the lbs of solids feed to the belt press (5 points)  
(b) Calculate lbs cake hauled by the trucks over the three days (5 points)  
(c) Calculate the lbs solids in the cake hauled (given the solids % of biosolids) (5 points)  
(d) Calculate percent solids recovery (lbs solids produced per lbs feed solids) (5 points)  
Solution:  
(a) lbs solids feed to belt press:  
gal  
min  
lbs  
gal  
lbs solids  
lb feed sludge  
min  
day  
lbs solids  
day  
80  
8.34  
0.025  
(1060)  
= 10,008  
(b) lbs cake hauled by truck over three days:  
yd3 sludge  
truck  
trucks  
three days  
ft3  
yd3  
lbs cake  
ft3 sludge  
lbs cake  
three days  
13.3  
5  
27  
70  
= 125,685  
(c) lbs solids in the cake hauled:  
lbs cake  
three days  
lbs solids  
lbs cake  
lbs solids  
three days  
125,685  
0.22  
= 27,651  
(d) Percent solids recovery:  
lbs solids  
27,651  
lbs solids in cake  
three days  
lbs solids  
=
=
= 0.92 or 92%  
lbs solids fed to belt press  
10,0083  
three days  
41. Lab data from your 100,000 gallon primary anaerobic digester, which receives primary sludge only,  
is shown below. Using this data :  
(a) Calculate the average volatile solids reduction. Compare your calculated value to generally  
accepted ranges for a healthy anaerobic digester. Comment.  
(b) Compare the other data to expected ranges.  
(c) Is this digester experiencing an operational problem ? If so, what is the problem. Name three  
steps that may be taken to mitigate the problem.  
(d) Should slake lime be added ? Why or why not ?  
Data  
CO2  
Date  
pH  
Alkalinity  
(mg/l)  
Volatile  
acids  
Raw  
Raw  
Digested  
Sludge  
Sludge  
Sludge  
(mg/l)  
280  
(TS%)  
5.4  
(VS%)  
65.5  
(VS%)  
56  
9/02  
9/09  
9/16  
9/17  
7.10  
7.00  
6.90  
6.85  
3,200  
3,020  
2,800  
2,720  
35.5  
36.0  
37.7  
38.2  
320  
5.0  
66.7  
53.8  
54.2  
400  
4.9  
65.9  
450  
42. Two sludges are blended together as follows: 8,000 cu. ft primary sludge at 4.90% solids and  
23,000 gal. secondary sludge at 5.20% solids. What is the combined solids concentration?  
Solution:  
Correct Answer:  
C1 V1 +C2 V2 = C3 V3  
7.48gal  
4.98,000 ft3 ∗  
+5.223,000  
ft3  
C1 V1 +C2 V2  
V3  
C1 V1 +C2 V2  
V1 +V2  
=C3 =  
=
=
=
80007.48+23,000  
4.98% Correct Answer  
448  
Chapter 22. Water Math  
Incorrect Answer #1  
C1 V1 +C2 V2 = C3 V3  
C1 V1 +C2 V2  
C1 V1 +C2 V2  
V1 +V2  
4.98,000ft3 +5.223,000  
=C3 =  
=
=
= 5.12% −  
V3  
8000+23,000  
Incorrect Answer#1  
Incorrect Answer #2  
6.23%  
Incorrect Answer #3  
3.84%  
43. You have two 110 foot diameter primary clarifiers. The raw wastewater flow of 1.5 MGD is divided  
equally between the these two basins. Raw settled primary sludge is pumped continuously from  
both clarifiers for thickening in a single 40 foot diameter gravity thickener. The average suspended  
solids are 210 mg/L for the raw influent flow and 85 mg/L for the primary effluent. Calculate the  
average solids loading on the thickener in pounds/ft2/day Ans. 1.25lbs/ft2/day  
44. A belt filter press is used 10 hrs per day to dewater 80 GPM digester sludge feed with 2.5% solids.  
45. A belt filter press is used 10 hrs per day to dewater 80 GPM digester sludge feed with 2.5% solids.  
The belt press produces 22% biosolids (bulk density - 70lbs/cu. ft) which is loaded into trucks for  
offsite disposal. Five truckloads of biosolids are produced every three days with each truckload  
averaging 13.3 cu. yd of material.  
(a) Calculate the lbs of solids feed to the belt press (5 points)  
(b) Calculate lbs cake hauled by the trucks over the three days (5 points)  
(c) Calculate the lbs solids in the cake hauled (given the solids % of biosolids) (5 points)  
(d) Calculate percent solids recovery (lbs solids produced per lbs feed solids) (5 points)  
(e) Two sludges are blended together as follows: 8,000 cu. ft primary sludge at 4.90% solids and  
23,000 gal. secondary sludge at 5.20% solids. What is the combined solids concentration?  
a. 3.84  
b. 6.23  
c. 5.12  
*d. 4.98  
46. 325 GPM of WAS with 7,000 mg/l TSS is thickened in a DAFT. Assuming an air:solids ratio of  
0.05 lb of air for each lb of feed solids, The SCFM (standard cubic feed per minute) air required to  
thicken the WAS given 1 SCFM = 0.075 lb is:  
*a. 12.6 SCFM  
b. 0.126 SCFM  
c. 12,600 SCFM  
d. 1.26 SCFM  
47. When you spread sludge on agricultural land, the annual application rate of cadmium in the sludge  
should be less than 2 lbs/acre/year If your sludge contains 30 mg cadmium per kilogram of solids  
and your plant produces 950,000 lbs per year of dry solids, how many acres do you need?  
a. 3 acres  
*b. 14 acres  
c. 19 acres  
d. 27 acres  
48. 42,000 gallons of 6% sludge containing 67% volatile matter is pumped to the digester. The di-  
22.27 Dewatering math problems  
449  
gester reduces the volatile matter by 52%. What volume of sludge in gallons containing 5% solids  
remains after digestion? [Ans: 32,841 gal]  
49. An Imhoff cone result is 5.5 ml/l. Approximately how many gallons of primary sludge will need to  
be pumped if the flow is 0.7 MGD? [3,850 gallons]  
50. A sludge digester equipped with a floating cover is 31 ft. inside diameter. The corbels are 16 feet  
above the floor and the top of the wall is 20 feet above the floor. The digester currently has a sludge  
depth of 16.5 ft. How many gallons of sludge are needed to displace the cover 1 foot? [5,643  
galllons]  
51. 10,000 gallons of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). 50% of  
the VS are destroyed, creating 10 ft.3 of gas per lbs. of VS destroyed. How much gas is produced  
each day? [Ans. 11,676 cu. ft/day]  
52. Two sludges are blended together as follows: 15,000 gal. primary sludge at 4.1% solids. 28,000 gal.  
secondary sludge at 1.3% solids. What is the combined solids concentration? [Ans. 2.28%] If the  
primary sludge is 68% VS and the secondary sludge is 63% VS, how many pounds of VS are in the  
combined sludge? [Ans. 5400.3 lbs VS]  
53. 12,000 gallons of 1.8% sludge is pumped to a thickener, and is thickened to 3800 gallons of 4.9%  
solids. The supernatant is returned back to the head of the STP for treatment. What is the volume in  
gallons and solids concentration in ppm of the supernatant?[Ans. 24,980 gallons]  
54. How many pounds of solids are pumped to a digester each day if the digester receives 10,000 gpd  
of sludge at 5% solids concentration?  
Solution:  
lbs TS 10,000gal sludge (8.340.05lbsTS)  
= 4,170  
lbs TS  
day  
=
day  
day  
gal sludge  
55. Calculate the % VS reduction in a digester given the volatile solids content of the influent sludge to  
the digester is 70% and the volatile solids content of the sludge leaving the digester is 52.5%  
0.70.525  
Solution: Digester VS reduction(%) =  
100 = 53%  
0.70.70.525  
56. Calculate the VS loading to the digester in lbs/day if 10,000 gallons of sludge containing 5% TS  
with and average VS content of 78%  
Solution:  
Digester VS loading (lbs/day)  
10,000 gallons sludge 8.34lbs sludge 0.050.78lbsVS  
=
= 3,253lbs sludge per day  
day  
gal  
lb sludge  
57. The sludge feed to a digester is 12,000 ft3/day. The sludge contains 4.5% total solids with 70%  
volatile solids. What is the digester loading in lbs VSS/day per ft3 if the digester diameter is 100 ft  
and a working sludge level of 20 ft.  
Solution:  
lbs VS  
Digester loading  
day  
Digester volatile solids loading Rate =  
Digester volume(V)ft3  
450  
Chapter 22. Water Math  
ft3 sludge  
day  
gal  
ft3  
lbsVS  
gal sludge  
12,000  
7.48  
(8.340.0450.70)  
lbs VS  
= 0.15  
=
π
dayft3  
(
1002 20)ft3  
4
58. How much gas is produced in the above digester in ft3/day if the digested sludge contains 2.5%  
total solids of which 59% is volatile solids and the gas production rate is 14 ft3/lb VS destroyed?  
Solution:  
0.70.59  
Digester VS reduction(%) =  
100 = 38.3%  
0.700.700.59  
lbs VS reduction 3153lbs VS feed 0.383 VS reduction  
lbs VS reduction  
day  
=
= 1,208  
day  
day  
ft3gas produced  
day  
lbs VS reduced 14 ft3 gas produced  
ft3 digester gas produced  
= 1208  
= 16,912  
day  
lb VS reduced  
day  
59. The sludge feed to a digester is 80,000 gal/day. The sludge contains 4.5% total solids with 75%  
volatile solids. If 50% of VS are reduced in the digester:  
(a) Find lbs VS destroyed per 1000 gal of digester capacity per day if the digester radius is 55 ft  
with an operating sludge level of 25 ft.(5 points)  
Solution:  
gal  
ft3  
Digester Volume: (π 552 25) ft3 7.48  
= 1,776,220gallons = 1,776.2 (1000 gallons)  
lbs  
gal  
lbsVS  
lb  
lbs VS reduction  
0.5  
lb VS  
80,000gal 8.34  
(0.0450.75)  
lbs VS reduction  
day  
=
day  
lbs VS reduction  
day  
= 11,259  
lbs VS reduction  
day  
1,776.2 (1000 gallons)  
11,259  
lbs VS reduction  
1000 digester capacity  
lbs VS reduction  
1000gallons digester volume  
=
= 6.3  
(b) What is the digester gas production in Btu/day? Assume 14 cu. ft digester gas per lb of VS  
destroyed and a 650 Btu/cu. ft heating value for the digester gas produced (5 points)  
Solution:  
ft3gas produced  
day  
lbs VS reduced 14 ft3 gas produced  
ft3 digester gas produced  
day  
= 11,259  
= 157,626  
day  
lb VS reduced  
BTU produced  
ft3 gas produced 650BTU gas produced  
BTU produced  
= 157,626  
= 102,456,900  
day  
day  
ft3gas  
day  
60. Two sludges are blended together as follows: 15,000 gal. primary sludge at 4.1% solids. 28,000 gal.  
secondary sludge at 1.3% solids. What is the combined solids concentration? If the primary sludge  
is 68% VS and the secondary sludge is 63% VS, what is the VS concentration (%) in the combined  
sludge?  
22.27 Dewatering math problems  
451  
Solution:  
Combined solids concentration: C1 V1 +C2 V2 = C3 V3  
C1 V1 +C2 V2 C1 V1 +C2 V2  
4.115,000+1.328,000  
= 2.28%  
=C3 =  
=
=
V3  
V1 +V2  
15,000+28,000  
Lbs of VS in combined sludge: CVS1 V1 +CVS2 V2 = CVS3 V3  
CVS1 V1 +CVS2 V2 CVS1 V1 +CVS1 V2  
=CVS3  
=CVS3  
=
=
=
V3  
V1 +V2  
4.115,0000.68+1.328,0000.63  
15,000+28,000  
= 1.50%  
452  
Chapter 22. Water Math  
61. 10,000 gallons of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). 50% of  
the VS are destroyed, creating 15 ft3 of gas per lbs. of VS destroyed. How much gas is produced  
each day?  
lbs  
gal  
lbsVS 0.5lbs VS reduction  
10,000gal 8.34  
(0.040.7)  
lbs VS reduction  
day  
lbs VS reduction  
day  
lb  
lb VS  
=
= 1,168  
day  
ft3gas produced  
day  
lbs VS reduced 15 ft3 gas produced  
ft3 digester gas produced  
day  
= 1,168  
= 17,520  
day  
lb VS reduced  
62. Two sludges are blended together as follows: 60,000 gal per day. primary sludge at 4.5% solids and  
28,000 gal secondary sludge at 5% solids.  
(a) What is the combined solids concentration? (3 points)  
60,0004.5+28,0005  
= 4.66%  
88,000  
(b) How many lbs of solids (per day) are in the combined sludge (3 points)  
88,000gal 8.340.0466 = 34,200lbs  
(c) If the primary sludge is 68% VS and the secondary sludge is 83% VS, how many pounds of  
VS are in the combined sludge? (3 points)  
8.340.0450.68lbs VS  
8.340.050.83lbs VS  
60,000gal∗  
+28,000gal∗  
= 25,003lbs VS  
gal  
gal  
(d) What is the digester feed VS%? (3 points)  
25,003lbs VS  
34,200lbsTS  
= 73.1%  
(e) If the digester is 120’ diameter with a liquid height of 20’, what is the VS loading in lbs  
VS/ft3-day (3 points)  
25,003lbs VS  
0.7851202 20  
0.11lbs VS  
dayft3  
=
(f) If the digested sludge has 52% VS, calculate the digester VS reduction percent (3 points)  
22.27 Dewatering math problems  
453  
0.731.52  
Digester VS reduction(%) =  
100 = 60.1%  
0.7310.7310.52  
(g) What is the gas production (ft3/day) if the digester produces 14 ft3 gas/lb VS destroyed (3  
points)  
lbs VSdestroyed 14 ft3 gas produced  
25,003*0.601  
= 210,375ft3gas  
lb VS destroyed  
(h) How many belt presses are needed to keep up with the digested sludge flow if the belt presses  
can be operated at a maximum flow of 50 GPM (3 points)  
gal sludge  
day  
day  
1440min  
88,000  
= 61GPM  
@50 GPM per press - 2 BFP required  
(i) If the digested sludge feed to the belt filter presses is 2.6% and assuming the belt press feed  
solids capture is 90%. How many lbs of solids (dry) are produced by the BFP (4 points)  
gal  
day  
lbs solids feed  
day  
lbs solids captured  
lbs solids feed  
88,000  
8.340.026  
0.9  
= 17,174lbs solids per day  
(j) If the BFP produces a 20% cake, how many wet tons cake produced per day (4 points)  
17,174lbs solids produced  
100lbs cake  
20lbs solids produced 2,000lbs  
tons  
42.9tons  
day  
=
day  
(k) If the cake density is 68 lbs/cu. ft, how much time will it take to fill a 30 cu. yd bin (3 points)  
(42.92000)lbs cake  
day  
59.583  
(Ans. 15.4 hours)  
=
min  
day  
1440min lbs cake  
59.583  
min∗  
ft3  
0.876 ft3  
=
lbs cake  
68lbs cake  
min  
27 ft3  
yd3  
min  
hrs  
30yd3 ∗  
0.876ft3 60min  
= 15.4hrs  
454  
Chapter 22. Water Math  
(l) What will be the cost of hauling dewatered cake per day @ $65 per ton cake (2 points)  
tons $65  
day ton  
$2,789  
day  
42.9  
=
63. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS  
with and average VS content of 76% is fed to the digester  
Solution:  
25,000 Gal 8.34 lbs sludge (0.0450.76) lbs VS feed  
=
7,131 lbs VS  
day  
Gal  
lb sludge  
day  
64. Calculate the organic loading rate to two 320,000 gallon anaerobic digesters given: Primary sludge  
feed rate of 25,000 gallons with 6.2% TS containing 73% solids and SG of 1.03. Secondary sludge  
feed rate of 30,500 gallons with 3.8% TS containing 77% solids (7 points) (Answer: 0.2lbs VS/day-  
ft3 )  
65. 10,000 gallons/day of sludge is pumped to an anaerobic digester/day at 4% solids (70% VS). If  
50% of the VS is destroyed, how many lbs of VS is destroyed per day?  
Solution:  
10,000 Gal 8.34 lbs sludge 0.040.7 lbs VS feed 0.5 lbs VS destroyed  
1,168lbs VS destroyed  
=
day  
Gal  
lb sludge  
lbs VS feed  
day  
66. Two sludges are blended together as follows: 8,000 cu. ft primary sludge at 4.90% solids and  
23,000 gal. secondary sludge at 5.20% solids. What is the combined solids concentration?  
a. 3.84  
b. 6.23  
c. 5.12  
*d. 4.98  
67. Calculate the VS loading to the digester in lbs/day if 10,000 cu. ft of sludge containing 4.8% TS  
with and an average VS content of 82% (Ans: 24,554)  
68. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS  
with and average VS content of 76% is fed to the digester (Ans: 7,131)  
69. Calculate the % VS reduction in a digester given the volatile solids content of the feed sludge to the  
digester is 76% and the volatile solids content of the sludge leaving the digester is 55% (Ans: 61.4)  
70. If an anaerobic digester receives sludge with VS of 76% and discharges digested sludge with a 56%  
VS. Its VS reduction is:  
a. 36%  
b. 48%  
*c. 60%  
d. 89%  
71. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS  
with and average VS content of 76% is fed to the digester  
Correct Answer(s): a. 7131.0  
72. Calculate the organic loading rate to two 320,000 gallon anaerobic digesters given: Primary sludge  
feed rate of 25,000 gallons with 6.2% TS containing 73% solids and SG of 1.03. Secondary sludge  
feed rate of 30,500 gallons with 3.8% TS containing 77% solids (7 points) (Answer: 0.2lbs VS/day-  
ft3 )  
Correct Answer(s):  
73. Calculate the VS loading to the digester in lbs/day if 25,000 gallons of sludge containing 4.5% TS  
with and average VS content of 76% is fed to the digester  
*a. 7,131 lbs/day  
22.27 Dewatering math problems  
455  
b. 17,432 lbs/day  
c. 19,256 lbs/day  
d. 26,244 lbs/day  
74. What is the digester efficiency if the volatile solids concentration entering an anaerobic digester is  
70% and the volatile solids concentration leaving the digester is 50%?  
*a. 29%  
b. 40%  
c. 57%  
d. 71%  
75. You are operating a 1 MGD trickling filter plant with a single stage anaerobic digester You pump  
about 3,000 gpd of sludge to the fixed cover digester. The supernatant has a BOD of 950 mg/L The  
BOD loading coming from the supernatant in pounds per day is  
a. 12  
*b. 24  
c. 36  
d. 48  
76. How many pounds of solids are pumped to a digester each day, if the digester receives 10,000 gpd  
of sludge at 5% density?  
a. 417 lbs/day  
b. 1,497 lbs/day  
c. 2,337 lbs/day  
d. 3,273 lbs/day  
*e. 4,170 lbs/day  
77. If a digester loading is 0.05 lbs VSS/ft3/Day, how large must the digester be if it receives 1,000 lbs.  
of suspended solids per day with a volatile percentage of 75%?  
a. 3,750 cu.ft  
b. 10,000 cu.ft  
*c. 15,000 cu.ft  
d. 26,666 cu.ft  
e. 37,500 cu.ft  
78. What is the digester efficiency if the volatile solids concentration entering an anaerobic digester is  
70% and the volatile solids concentration leaving the digester is 50%?  
*a. 29%  
b. 40%  
c. 57%  
d. 71%  
79. You are operating a 1 MGD trickling filter plant with a single stage anaerobic digester You pump  
about 3,000 gpd of sludge to the fixed cover digester. The supernatant has a BOD of 950 mg/L The  
BOD loading coming from the supernatant in pounds per day is  
a. 12  
*b. 24  
c. 36  
d. 48  
80. How many pounds of solids are pumped to a digester each day, if the digester receives 10,000 gpd  
of sludge at 5% density?  
456  
Chapter 22. Water Math  
a. 417 lbs/day  
b. 1,497 lbs/day  
c. 2,337 lbs/day  
d. 3,273 lbs/day  
*e. 4,170 lbs/day  
81. If a digester loading is 0.05 lbs VSS/ft3/Day, how large must the digester be if it receives 1,000 lbs.  
of suspended solids per day with a volatile percentage of 75%?  
a. 3,750 cu.ft  
b. 10,000 cu.ft  
*c. 15,000 cu.ft  
d. 26,666 cu.ft  
e. 37,500 cu.ft  
Disinfection Math Problems  
1. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/l at a  
5.0 MGD flow.  
Solution:  
lbs/day = conc.(mg/l)flow(MGD)8.34  
lbs/day = 1258.34 = 500.4lbs/day  
2. What is the chlorine demand if the chlorine dosage is 15 mg/L and the residual is 3 mg/l? Solution:  
Chlorine dosage = chlorine demand + chlorine residual  
=chlorine demand = chlorine dosagechlorine residual = 153 = 12mg/l  
3. If 80 pounds of chlorine are applied each day to a flow of 1.5 MGD, what is the dosage in mg/l?  
Solution:  
Applying the pounds formula:  
lbs/day = conc.(mg/l)flow(MGD)8.34  
lbs/day  
80  
=conc.(mg/l) =  
=
= 6.4mg/l  
1.58.34  
flow(MGD)8.34  
4. How many pounds per day of chlorine will be required to disinfect a secondary effluent flow of  
1.68 MGD if the chlorine demand is found to be 8.5 mg/l and a residual of 3 mg/l is desired? Chlo-  
rine dosage = chlorine demand + chlorine residual  
chlorine dosage = 8.5+3 = 11.5mg/l  
lbs/day = conc.(mg/l)flow(MGD)8.34 = 1.6811.58.34 = 161.2lbs/day  
5. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a flow of 2.8  
MGD, how many pounds per day should the chlorinator be set to deliver.  
Solution:  
Chlorine dosage = chlorine demand + chlorine residual  
chlorine dosage = 4.8+0.75 = 5.55mg/l  
To calculate pounds per day, applying the pounds formula:  
lbs/day = conc.(mg/l)flow(MGD)8.34 = 2.85.558.34 = 129.6lbs/day  
6. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
22.27 Dewatering math problems  
457  
d. 8.3 mg/L  
Solution:  
Chlorine dosage(lbs/day) = conc.(mg/l)flow(MGD)8.34  
lbs/day  
75  
=chlorine dosage conc.(mg/l) =  
=
= 7.5mg/l  
1.28.34  
flow(MGD)8.34  
Chlorine dosage = chlorine demand + chlorine residual  
=chlorine demand = chlorine dosagechlorine residual = 7.52.6 = 4.9mg/l  
7. Experience has shown that a minimum dosage of 24 mg/l is necessary in order to disinfect a  
wastewater effluent and leave a residual of 1.0 mg/l. How many pounds of chlorine must be fed  
at this dosage to a flow of 0.5 MGD?  
Solution:  
Chlorine dosage(lbs/day) = conc.(mg/l)flow(MGD)8.34 = 240.58.34 = 100lbs/day  
8. 25 lbs/day of chlorine is being applied to a wastewater effluent flow of 250,000 gpd. Calculate the  
chlorine dosage in mg/l.  
Solution:  
lbs/day = conc.(mg/l)flow(MGD)8.34  
lbs/day  
25  
=conc.(mg/l) =  
=
= 12mg/l  
0.258.34  
flow(MGD)8.34  
9. You wish to dose the influent channel at 5 mg/l chlorine to help control odors. The flow is 11.5  
MGD.. How many pounds of chlorine must be fed each day? Is it necessary to maintain a chlorine  
residual to control odors?  
Solution:  
lbs/day = conc.(mg/l)flow(MGD)8.34 = 511.58.34 = 480lbs/day  
No. Chlorine residual is for disinfection only. Odor control would not require a residual  
10. Jar testing shows that the chlorine demand of an effluent is 12.5 mg/l. In order to assure disinfec-  
tion, a residual of 1.0 mg/l is required. How many pounds of chlorine must be fed per 1MGD to  
assure disinfection?  
Solution: chlorine dosage = chlorine demand + chlorine residual  
=chlorine dosage = (12.5 + 1)mg/l = 13.5mg/l  
lbs/day = 13.5(mg/l)1(MGD)8.34 = 112.6lbs/day  
11. What is the chlorine dosage if the chlorinator is feeding 120 lbs/day and the average daily flow is  
3.5 MGD? What is the chlorine demand if the residual is 1.3 mg/l?  
Solution:  
a. Chlorine dosage(lbs/day) = conc.(mg/l)flow(MGD)8.34  
lbs/day  
flow(MGD)8.34  
120  
3.58.34  
=chlorine dosage conc.(mg/l) =  
=
= 4.1mg/l  
b. Chlorine dosage = chlorine demand + chlorine residual  
=chlorine demand = chlorine dosagechlorine residual = 4.11.3 = 2.8mg/l  
12. Experience has shown that a minimum dosage of 24 mg/l is necessary in order to disinfect a  
wastewater effluent and leave a residual of 1.0 mg/l. How many pounds of chlorine must be fed  
at this dosage to a flow of 0.5 MGD?  
Solution:  
Chlorine dosage(lbs/day) = conc.(mg/l)flow(MGD)8.34 = 240.58.34 = 100lbs/day  
13. 25 lbs/day of chlorine is being applied to a wastewater effluent flow of 250,000 gpd. Calculate the  
chlorine dosage in mg/l.  
Solution:  
lbs/day = conc.(mg/l)flow(MGD)8.34  
lbs/day  
25  
=conc.(mg/l) =  
=
= 12mg/l  
0.258.34  
flow(MGD)8.34  
458  
Chapter 22. Water Math  
14. You wish to dose the influent channel at 5 mg/l chlorine to help control odors. The flow is 11.5  
MGD.. How many pounds of chlorine must be fed each day? Is it necessary to maintain a chlorine  
residual to control odors?  
Solution:  
lbs/day = conc.(mg/l)flow(MGD)8.34 = 511.58.34 = 480lbs/day  
No. Chlorine residual is for disinfection only. Odor control would not require a residual  
15. Jar testing shows that the chlorine demand of an effluent is 12.5 mg/l. In order to assure disinfec-  
tion, a residual of 1.0 mg/l is required. How many pounds of chlorine must be fed per 1MGD to  
assure disinfection?  
Solution: chlorine dosage = chlorine demand + chlorine residual  
=chlorine dosage = (12.5 + 1)mg/l = 13.5mg/l  
lbs/day = 13.5(mg/l)1(MGD)8.34 = 112.6lbs/day  
16. What is the chlorine dosage if the chlorinator is feeding 120 lbs/day and the average daily flow is  
3.5 MGD? What is the chlorine demand if the residual is 1.3 mg/l?  
Solution:  
a. Chlorine dosage(lbs/day) = conc.(mg/l)flow(MGD)8.34  
lbs/day  
flow(MGD)8.34  
120  
3.58.34  
=chlorine dosage conc.(mg/l) =  
=
= 4.1mg/l  
b. Chlorine dosage = chlorine demand + chlorine residual  
=chlorine demand = chlorine dosagechlorine residual = 4.11.3 = 2.8mg/l  
17. A 2.5 MGD secondary flow is disinfected by the application of 320 lbs of chlorine per day. This  
dose provides a chemical residual of 2.1 mg/l. There is a need to switch to the use of sodium  
hypochlorite which has a 12.5% available chlorine, SG of 1.2 and a cost of $0.60 per gallon. Chlo-  
rine costs $0.28/lb.  
Calculate: 1) The chlorine demand, and 2) Cost difference ($ per day) between chlorine and  
sodium hypochlorite  
Dosage = Demand + Residual  
Dosage:  
mg  
l
320lbs chlorine  
= 2.5MGD8.34x  
day  
xmg  
=
= 15.34mg  
320  
2.58.34  
l
l
mg  
l
Chlorine Demand = Dosage - Residual = 15.34 - 2.1 = 13.24  
$320 $0.28  
Cost per day to use chlorine:  
= $89.60  
lb  
lb  
To calculate the hypochlorite we need to determine the gallons per day of bleach required.  
bleach  
day  
320lbs  
= xgal  
8.341.2 per  
bleach 0.125lbs  
chlorine  
day  
lbs bleach  
chlorine  
gal  
lb bleach  
bleach  
day  
bleach  
xgal  
=
= 256gal  
320  
8.341.20.125  
$0.60  
day  
256gal  
= $153.48 Cost difference $153.48 - $89.60 = $63.88  
gal bleach  
bleach  
day  
22.27 Dewatering math problems  
459  
18. Three hundred pounds (300 lbs) per day of chlorine (cost = $0.32/lb) are being used to disinfect a  
secondary effluent flow of 6.5 MGD. Due to safety concerns, the substitution of sodium hypochlo-  
rite (12.5% available chlorine, 10.0 lbs/gal and a cost of $0.55/gal) for the chlorine is being consid-  
ered at this plant. What is the cost difference per day between the chlorine and sodium hypochlo-  
rite? (Ans:$36/year)  
19. The operator at a 1.5 MGD conventional activated sludge plant is considering using either HTH or  
sodium hypochlorite as an alternative to chlorine gas. Currently chlorine is being dosed at 15 mg/L  
in order to achieve a residual of 3.0 mg/L. Using the data provided below calculate the daily cost  
for chlorine, HTH, and sodium hypochlorite (NaOCl) (Sp.Gravity 1.21).  
Chlorine 0.15 $/lb  
HTH (70% available chlorine) 0.25 $/lb  
NaOCl (15% available chlorine) 0.35 $/gal  
SOLUTION  
lbs chlorine required:  
15mg chlorine  
1.5MG 8.34lbs  
188lbs chlorine  
=
day  
gallon  
l
day  
Daily cost if chlorine is used:  
$0.15  
chlorine  
188lbs chlorinelb  
= $28.20  
Daily cost if HTH is used:  
lb HTH  
$0.25  
188lbs chlorine0.7lb  
= $67.14  
chlorine lb chlorine  
Daily cost if NaOCl is used:  
188lbs chlorine0.15lb  
gal NaOCl  
chlorine 8.341.21lbs NaOCl gal NaOCl  
lb NaOCl  
$0.35  
= $43.47  
460  
Chapter 22. Water Math  
20. A water storage tank is 30 feet in diameter and has a water depth of 18.5 feet. It is desired to super-  
chlorinate this tank with 30 ppm of chlorine, how many pounds of HTH will be required (HTH has  
70% available chlorine)?  
Tank Volume:  
0.785302 18.5 ft3 7.48gfat l = 97,765gal  
3
lbs HTH required:  
30lbs chlorine  
8.34lbs water  
lb HTH  
97,765gal water 0.70lb  
= 35lbs HTH  
chlroine  
1,000,000lbs water  
gal water  
21. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a flow of 2.8  
MGD, how many pounds per day should the chlorinator be set to deliver.  
*a. 130  
b. 116  
c. 112  
d. 18  
e. 16  
22. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
23. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5  
mg/L, approximately how many pounds of chlorine per day will be required to treat a flow of 5.0  
MGD?  
a. 21 lbs  
b. 792 lbs  
c. 813 lbs  
*d. 834 lbs  
24. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
25. If 25 Ibs/day of chlorine is being applied to a wastewater effluent flow of 250,000 gpd. Calculate  
the chlorine demand in mg/L if the chlorine residue is 1.2 mg/L.  
a. 2.5 mg/L  
*b. 10.8 mg/L  
c. 15.1 mg/L  
d. 6.3 mg/L  
26. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/L at  
6.0 MGD flow  
a. 60 lbs/day  
b. 450 lbs/day  
22.27 Dewatering math problems  
461  
c. 500 lbs/day  
*d. 600 lbs/day  
e. 6,700 lbs/day  
27. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
28. Chlorine is being fed at the rate of 200 pounds per day. Plant flow is 4 MGD. The chlorine residual  
is measured and found to be 3 mg/L Calculate chlorine demand.  
a. 3 mg/L  
*b. 2 mg/L  
c. 16 mg/L  
d. 9 mg/L  
29. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a flow of 2.8  
MGD, how many pounds per day should the chlorinator be set to deliver.  
*a. 130  
b. 116  
c. 112  
d. 18  
e. 16  
30. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
31. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5  
mg/L, approximately how many pounds of chlorine per day will be required to treat a flow of 5.0  
MGD?  
a. 21 lbs  
b. 792 lbs  
c. 813 lbs  
*d. 834 lbs  
32. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
33. If 25 Ibs/day of chlorine is being applied to a wastewater effluent flow of 250,000 gpd. Calculate  
the chlorine demand in mg/L if the chlorine residue is 1.2 mg/L.  
a. 2.5 mg/L  
*b. 10.8 mg/L  
462  
Chapter 22. Water Math  
c. 15.1 mg/L  
d. 6.3 mg/L  
34. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/L at  
6.0 MGD flow  
a. 60 lbs/day  
b. 450 lbs/day  
c. 500 lbs/day  
*d. 600 lbs/day  
e. 6,700 lbs/day  
35. What is the chlorine residual of a 2.5 MGD secondary effluent stream if its chlorine demand is 9  
mg/l and is treated with 300 lbs/day chlorine? Ans: 5.4  
36. How many lbs per day of chlorine is required to maintain a dosage of 12 mg/l for a 5 MGD flow?  
Ans: 500.4  
37. What is the chlorine residual of a 2.5 MGD secondary effluent stream if its chlorine demand is 9  
mg/l and is treated with 300 lbs/day chlorine? Ans: 5.4  
38. How many lbs per day of chlorine is required to maintain a dosage of 12 mg/l for a 5 MGD flow?  
Ans: 500.4  
39. What is the chlorine residual of a 2.5 MGD secondary effluent Ans: 5.4  
40. How many lbs per day of chlorine is required to maintain a dosage of 12 mg/l for a 5 MGD flow?  
Ans: 500.4  
41. Experience has shown that a minimum dosage of 24 mg/l is necessary in order to disinfect a  
wastewater effluent and leave a residual of 1.0 mg/l. How many pounds of chlorine must be fed  
at this dosage to a flow of 0.5 MGD? Ans: 100.0  
42. 25 lbs/day of chlorine is being applied to a wastewater effluent ow of 250,000 gpd. Calculate the  
chlorine dosage in mg/l. Ans: 12.0  
43. You wish to dose the in influent channel at 5 mg/l chlorine to help control odors. The flow is 11.5  
MGD. How many pounds of chlorine must be fed each day? Is it necessary to maintain a chlorine  
residual to control odors? Ans:480.0  
44. Jar testing shows that the chlorine demand of an effluent is 12.5 mg/l. In order to assure disinfec-  
tion, a residual of 1.0 mg/l is required. How many pounds of chlorine must be fed per 1MGD to  
assure disinfection? Ans: 112.6  
45. What is the chlorine dosage if the chlorinator is feeding 120 lbs/day and the average daily flow is  
3.5 MGD? Ans: 4.1  
46. What is the chlorine residual of a 2.5 MGD secondary effluent stream if its chlorine demand is 9  
mg/l and is treated with 300 lbs/day chlorine? Ans: 5.4  
47. How many lbs per day of chlorine is required to maintain a dosage of 12 mg/l for a 5 MGD flow?  
Ans: 500.4  
48. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
49. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
22.27 Dewatering math problems  
463  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
50. Chlorine is being fed at the rate of 200 pounds per day. Plant flow is 4 MGD. The chlorine residual  
is measured and found to be 3 mg/L Calculate chlorine demand.  
a. 3 mg/L  
*b. 2 mg/L  
c. 16 mg/L  
d. 9 mg/L  
51. Chlorine is being fed at the rate of 200 pounds per day. Plant flow is 4 MGD. The chlorine residual  
is measured and found to be 3 mg/L Calculate chlorine demand.  
a. 3 mg/L  
*b. 2 mg/L  
c. 16 mg/L  
d. 9 mg/L  
52. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a flow of 2.8  
MGD, how many pounds per day should the chlorinator be set to deliver.  
*a. 130  
b. 116  
c. 112  
d. 18  
e. 16  
53. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
54. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5  
mg/L, approximately how many pounds of chlorine per day will be required to treat a flow of 5.0  
MGD?  
a. 21 lbs  
b. 792 lbs  
c. 813 lbs  
*d. 834 lbs  
55. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
56. If 25 Ibs/day of chlorine is being applied to a wastewater effluent flow of 250,000 gpd. Calculate  
the chlorine demand in mg/L if the chlorine residue is 1.2 mg/L.  
a. 2.5 mg/L  
464  
Chapter 22. Water Math  
*b. 10.8 mg/L  
c. 15.1 mg/L  
d. 6.3 mg/L  
57. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/L at  
6.0 MGD flow  
a. 60 lbs/day  
b. 450 lbs/day  
c. 500 lbs/day  
*d. 600 lbs/day  
e. 6,700 lbs/day  
58. How many pounds of chlorine will be used in one day if the flow is 700,000 gpd and a uniform  
dose of 12 mg/L is applied?  
*a. 7 lbs  
b. 15 lbs  
c. 22 lbs  
d. 26 lbs  
59. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5  
mg/L, approximately how many pounds of chlorine per day will be required to treat a flow of 50  
MGD?  
a. 21 lbs  
b. 792 lbs  
c. 813 lbs  
*d. 834 lbs  
60. How many pounds of chlorine gas is necessary to treat 4,000,000 gallons of wastewater at a dosage  
of 2 mg/L?  
a. 61 lbs  
b. 65 lbs  
*c. 67 lbs  
d. 69 lbs  
61. If you need a chlorine residual of 1 mg/l , how many pounds of chlorine must be applied each day  
if the fl ow is 2.5 MGD and the chlorine demand is l5 mg/l?  
a. 291 .pounds/day  
b. 312 pounds/day  
*c. 334 pounds/day  
d. 419 pounds/day  
e. 516 pounds/day  
62. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
63. Chlorine is being fed at the rate of 200 pounds per day. Plant flow is 4 MGD. The chlorine residual  
is measured and found to be 3 mg/L Calculate chlorine demand.  
a. 3 mg/L  
22.27 Dewatering math problems  
465  
*b. 2 mg/L  
c. 16 mg/L  
d. 9 mg/L  
64. The chlorine demand is 4.8 mg/l and a chlorine residual is 0.75 mg/l is desired. For a flow of 2.8  
MGD, how many pounds per day should the chlorinator be set to deliver.  
*a. 130  
b. 116  
c. 112  
d. 18  
e. 16  
65. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
66. Assuming that a chlorine residual of 0.5 mg/L is being maintained and the chlorine demand is 19.5  
mg/L, approximately how many pounds of chlorine per day will be required to treat a flow of 5.0  
MGD?  
a. 21 lbs  
b. 792 lbs  
c. 813 lbs  
*d. 834 lbs  
67. Chlorine is being fed at the rate of 75 pounds per day. Plant flow is 1.2 MGD. The chlorine residual  
is measured and found to be 2.6 mg/L Calculate chlorine demand.  
*a. 4.9 mg/L  
b. 5.7 mg/L  
c. 7.5 mg/L  
d. 8.3 mg/L  
68. If 25 Ibs/day of chlorine is being applied to a wastewater effluent flow of 250,000 gpd. Calculate  
the chlorine demand in mg/L if the chlorine residue is 1.2 mg/L.  
a. 2.5 mg/L  
*b. 10.8 mg/L  
c. 15.1 mg/L  
d. 6.3 mg/L  
69. Calculate how many pounds per day of chlorine should be used to maintain a dosage of 12 mg/L at  
6.0 MGD flow  
a. 60 lbs/day  
b. 450 lbs/day  
c. 500 lbs/day  
*d. 600 lbs/day  
e. 6,700 lbs/day  
466  
Chapter 22. Water Math  
Chemical Dosing Math Problems  
1. Polymer is being added at 0.2 mg/l in order to achieve a 98% capture efficiency for a belt press.  
The feed to the belt press is 75 gallons per minute, containing 2.5% solids. Given the polymer costs  
$250 per gallon of 4.5% active polymer with a specific gravity of 1.08. What is the cost of polymer  
per dry ton of solids captured  
Solution:  
lbs polymer required:  
sludge  
lbs sludge  
0.2lbs polymer  
751440gal  
8.34  
day  
gal sludge 1,000,000lbs sludge  
polymer  
day  
= 0.1801lbs  
gallons polymer solution required:  
polmyer  
day  
polymer solution  
8.341.08lbs polymer solution  
lbs polymer  
polymer solution  
0.1801lbs  
= xgal  
0.045lb  
day  
gal polymer solution  
polymer solution  
day  
=0.444gal  
Polymer cost:  
0.444gal polymer soultion  
$250  
gallon polymer soultion  
day  
$111  
=
day  
Dry tons of solids captured:  
sludge  
0.98 lbs solids captured  
751440gal  
8.340.025 lbs solids  
ton solids  
2000lbs solids  
day  
gal sludge  
lbs solids  
11tons dry solids  
=
day  
Polymer cost per dry ton of solids captured:  
$111perday  
= $10.09  
11tons dry solids per day  
2. A 50 MGD flow is being treated with 20 mg/L ferric chloride. How many lbs of ferric chloride is  
required daily  
lbs ferric chloride required:  
8,340lbs ferric chloride  
20mg ferric chloride  
50MG 8.34lbs  
=
day  
gallon  
l
day  
3. If the ferric chloride solution used contains 40% dry ferric chloride with a specific gravity of 1.4,  
what is its required feed rate in GPM (7 points)  
Required FeCl3 feed (gal/min) to feed 8,340 lbs ferric chloride:  
8,340lbs FeCl  
xgal FeCl soltn. 8.341.4lbs FeCl soltn  
0.4lbs FeCl  
3
1440min  
day  
3
3
3
=
day  
minute  
gal FeCl soltn.  
lbs FeCl soltn.  
3
3
1.24gal  
8,340  
8.341.40.41440  
=x =  
=
min  
4. What is the daily ferric chloride dosing cost if the ferric chloride cost is $580/dry ton ferric chloride  
(5 points)  
Daily ferric chloride dosing cost:  
$2,419  
day  
8,340lbs FeCl  
ton  
=
2000lbs ton FeCl  
3
$580  
3
day  
22.27 Dewatering math problems  
467  
5. A 0.5% (based on dry weight) solution of polymer is being fed to a secondary effluent prior to sand  
filtration. It is desired to dose this effluent at 2.5 mg/L. Assuming an effluent flow of 3000 gpm, at  
what rate (gpm) should the polymer feed pump be set?  
· lbs polymer required:  
3000gallons 8.34 lbs e f fluent  
2.5 lbs polymer  
0.0626lbs polymer  
=
1,000,000lbs e f fluent  
min  
gallon  
min  
Required pumping rate to feed 0.0626 lbs polymer minute:  
0.0626 lbs polymer  
x
gallon polymer solution 8.34 lbs polymer solution  
0.005 lbs polymer  
lbs polymer solution  
=
min  
minute  
gallon polymer solution  
1.5 gallon  
x
gallon polymersolution  
0.0626  
8.340.005  
=
=
minute  
min  
6. How many pounds of dry polymer should be added to how many gallons of water to make enough  
2% polymer solution to dose a 10 MGD secondary effluent flow at 3.0 ppm of polymer?  
lbs polymer required:  
8.34lbs e f fluent 3mg polymer  
250.2lbs polymer  
10MGD∗  
=
gallon  
l
day  
Required pumping rate to feed 250.2 lbs polymer minute:  
250.2lbs polymer  
xgallon polymer solution 8.34lbs polymer solution  
0.02lbs polymer  
lbs ploymer solution  
=
day  
day  
gallon polymer solution  
1500gallon  
xgallon polymer solution  
day  
250.2  
8.340.02  
=
=
day  
This can also be done as follows:  
1500gallon  
250.2lbs polymer lbs ploymer solution  
gallon polymer solution  
=
8.34lbs polymer solution  
day  
0.02lbs polymer  
day  
So dissolve 250.2 lbs polymer in 1500 gallons  
250.2lbs polymer  
15008.34lbs polymer solution  
0.02lbs polymer  
= 2%polymer  
lb polymer solution  
Check:  
=
7. A 0.35% solution of polymer is being fed to a secondary effluent prior to sand filtration. The poly-  
mer feed pump is set to pump 0.5 gpm to an effluent flow of 4200 gpm. What is the polymer dose  
rate in ppm?  
Pounds polymer pumped:  
0.5gal PS  
0.0146lbs Polymer  
8.34lbs PS 0.0035lbs  
P
=
min  
gal PS  
lb PS  
min  
Polymer dose rate:  
lbs polymer  
e f fluent (ppm or mg ) =  
= 0.42ppm polymer  
0.0146lbs min  
6
8.344,200  
l
6
10 lbs  
10 lbs e f fluent  
1,000,000  
8. Liquid alum (49% alum, sp. gravity 1.32, $1.85/gal)) is being used to remove phosphorus from  
a 600,000 gpd activated sludge effluent. Two hundred milligrams per liter (200 mg/L) of alum,  
Al2(S04)3.14H20, is required to give adequate removal of the phosphorus in this effluent. Cal-  
culate the daily cost of liquid alum needed to remove phosphorus. [Formula Weights: Al =27,  
Al2(S04)3.14H20 =594]  
468  
Chapter 22. Water Math  
lbs alum required:  
200mg alum  
0.6MG 8.34lbs  
1001lbs alum  
=
day  
gallon  
l
day  
Required liquid alum feed (gal/day) to feed 1001 lbs alum:  
xgallon liquid alum 8.341.32lbs liquid alum  
1001lbs alum  
0.49lbs alum  
lbs liquid alum  
=
day  
minute  
gallon liquid alum  
1001  
8.341.320.49  
=x =  
= 186gal  
Daily cost of liquid alum to remove phosphorous:  
$344.10  
day  
186gal  
day  
$1.85  
gal  
=
22.27 Dewatering math problems  
469  
9. A 1.5% polymer solution (based on dry weight) is to be fed at the rate of 3.5 pp to a secondary  
effluent flow of 4.0 MGD. (a) Calculate the polymer pump feed rate (gallon per minute) necessary  
to dose this secondary effluent. (b) How many gallons per day of polymer solution will be required  
for an average flow of 3470 gpm?  
(a)  
Polymer required:  
116.8lbs polymer  
43.58.34 =  
day  
gal  
Feed rate ( ):  
min  
gal  
min  
116.8lbs polymer 100lbs polymer solution  
gal polymer solution  
= 0.65  
8.34lbs polymer solution 1440min  
day  
day  
1.5lbspolymer  
(b)  
Polymer required:  
3470gal  
145.9lbs polymer  
MG  
1,000,000gal  
1440min  
day  
MGD3.58.34 =  
min  
day  
Feed rate (mgainl ):  
gal  
min  
145.9lbs polymer 100lbs polymer solution  
gal polymer solution  
= 0.81  
8.34lbs polymer solution 1440min  
day  
day  
1.5lbspolymer  
10. Liquid alum (Sp. gravity 1.32, 49% Alum, $1.70 per gallon,) is being used to remove phosphorus  
from a 595,000 gal/day secondary. A dose of fourteen milligrams per liter (14.0 mg/L) of alu-  
minum is required to give adequate removal of the phosphorus in this effluent. Calculate the daily  
cost of liquid alum. (Note: Dry alum contains 9.4% aluminum).  
Aluminum Required:  
69.47lbs aluminum  
0.595148.34 =  
day  
$
Cost of alum (day ):  
gal alum solution  
(8.341.32)lbs alum solution  
69.47lbs aluminum  
100lbs alum  
100lbs alum solution  
49lbs alum  
$1.70  
=
gal alum solution  
day  
9.4lbs aluminum  
$232.91  
day  
11. Polymer solution (0.5% weight to weight) is being fed at the rate of 0.83 gpm to a secondary efflu-  
ent flow of 1950 gpm prior to sand filtration. Calculate the polymer dose in units of ppm.  
lbs polymer  
10 lbs e f fluent  
mg  
l
Polymer Dose -  
which is  
or ppm:  
6
polymer solution(PS) & polymer(P)  
2.13mg  
0.83gal PS  
8.34lbs PS 0.005lbs  
P
min  
=
polymer  
(8.341950)  
1,000,000  
min  
gal PS  
lbs PS  
6
10 lbs e f fluent  
l
12. A 4.5% (weight to weight basis) solution of polymer is being fed to a secondary effluent prior to  
sand filtration. It is desired to dose at 2.4 mg/L. Assuming an effluent flow of 6550 gpm, at what  
rate (gallons per minute) should the polymer feed pump be set?  
Polymer feed rate GPM:  
gal  
min  
lbs polymer 100lbs polymer solution  
gal polymer solution  
2.4  
65508.3410  
= 0.35  
8.34lbs polymer solution  
6
min  
4.5lbspolymer  
470  
Chapter 22. Water Math  
13. Polymer is being added at 0.3 mg/l in order to achieve a 92% capture efficiency for a belt press.  
The feed to the belt press is 100 gallons per minute, containing 2.5% solids. Given the polymer  
costs $460 per gallon of 4% active polymer with a specific gravity of 1.1. What is the cost of  
polymer per dry ton of solids captured (10 points)  
Solution:  
lbs polymer required:  
sludge  
lbs sludge  
0.3lbs polymer  
1001440gal  
8.34  
day  
gal sludge 1,000,000lbs sludge  
polymer  
day  
= 0.36lbs  
gallons polymer solution required:  
polmyer  
day  
polymer solution  
day  
8.341.1lbs polymer solution  
lbs polymer  
polymer solution  
0.36lbs  
= xgal  
0.04lb  
gal polymer solution  
polymer solution  
day  
=0.982gal  
Polymer cost:  
0.982gal polymer soultion  
$460  
gallon polymer soultion  
day  
$451.26  
day  
=
Dry tons of solids captured:  
sludge  
0.92 lbs solids captured  
1001440gal  
8.340.025 lbs solids  
ton solids  
2000lbs solids  
day  
gal sludge  
lbs solids  
13.81tons dry solids  
=
day  
Polymer cost per dry ton of solids captured:  
$451.26perday  
= $32.67  
13.81tons dry solids per day  
14. A flow of 5 MGD is being treated with 9.8 mg/l aluminum using liquid alum of 48% strength and  
SG of 1.32. Alum has 19% aluminum. If the liquid alum costs $1.62 per gallon, what is the cost per  
day  
Solution:  
lbs aluminum required:  
aluminum  
day  
aluminum  
= 408.7lbs  
day  
5MGD8.349.8lbs  
Alum needed to meet this dosing need:  
liquid alum  
day  
lbs liquid alum  
gal liquid alum  
408.7lbs  
0.19lbs  
= xgal  
8.34 1.32 per  
= 407gal  
0.48lb  
aluminum  
day  
lbs alum  
liquid alum  
aluminum  
lb alum  
liquid alum  
day  
liquid alum  
day  
=xgal  
=
408.7  
8.341.320.480.19  
liquid alum  
day  
Cost per day=407gal  
= $659.45  
gal liquid alum  
$1.62  
15. Prior to sand filtration, a secondary effluent flow of 5 MGD is dosed with 0.75% strength poly-  
22.27 Dewatering math problems  
471  
mer solution to achieve a dose of 1.5mg/l of polymer. a) What is the lbs of dry polymer per day  
necessary to treat this effluent, and b) What is the required GPM feed of the 0.75% polymer:  
Solution:  
Correct Answer:  
lbs polymer  
a) lbs of dry polymer required (lbs formula)=5MGD8.341.5 = 62.55  
day  
gal  
min  
day  
x
1440  
polymer  
day  
min  
b) Flow rate of 0.75% strength polymer = 62.55lbs  
=
8.347500  
gal  
1,000,000  
MG  
gal  
62.551,000,000  
= 0.7GPM Correct Answer  
14408.347,500  
=x  
=
min  
Incorrect Answer 1:  
a) lbs of dry polymer required (lbs formula)=5MGD8.341.50.75 = 46.9  
lbs polymer  
day  
gal  
min  
day  
x
1440  
polymer  
day  
min  
b) Flow rate of 0.75% strength polymer = 62.55lbs  
=
8.347500  
gal  
1,000,000  
MG  
gal  
min  
46.91,000,000  
=x  
=
= 0.5GPM Incorrect Answer#1  
14408.347,500  
Incorrect Answer #2:  
62.6GPM Incorrect Answer#2  
Incorrect Answer#3:  
12.6GPM Incorrect Answer#3  
16. If a chemical costs $30 per ton, how much will it cost per year to treat a flow of 15 MGD if the  
average dose is 18 mg/L?  
Solution:  
Tons of chemical required per year: (use lbs formula)  
h
i
mg  
l
days  
lbs  
ton  
411 tons  
year  
15 MGD18  
8.34 day 365 year 2000lbs  
=
Chemical cost:  
411 tons $30  
= $12,328 per year  
ton  
year  
17. Prior to sand filtration, a secondary effluent flow of 5 MGD is dosed with 0.75% strength polymer  
solution to achieve a dose of 1.5mg/l of polymer. What is the required GPM feed of the 0.75  
a. 12.6 GPM  
b. 62.6 GPM  
*c. 0.7 GPM  
d. 0.5 GPM  
Practice Problems - Pumping Power Requirements  
1. If a pump is operating at 2,200 gpm and 60 feet of head, what is the water horsepower? If the pump  
efficiency is 71%, what is the brake horsepower?  
2. The water horsepower of a pump is 10Hp and the brake horsepower output of the motor is 15.4Hp  
.
What is the efficiency of the pump?  
3. The water horsepower of a pump is 25Hp and the brake horsepower output of the motor is 48Hp  
What is the efficiency of the pump?  
.
4. The efficiency of a well pump is determined to be 75%. The efficiency of the motor is estimated at  
472  
Chapter 22. Water Math  
94%. What is the efficiency of the well?  
5. If a motor is 85% efficient and the output of the motor is determined to be 10 BHp, what is the  
electrical horsepower requirement of the motor?  
6. The water horsepower of a well with a submersible pump has been calculated at 8.2 WHp. The  
Output of the electric motor is measured as 10.3BHp. What is the efficiency of the pump?  
7. Water is being pumped from a reservoir to a storage tank on a hill. The elevation difference be-  
tween water levels is 1200 feet. Find the pump size required to fill the tank at a rate of 120 gpm.  
Express your answer in horsepower.  
8.  
A
25hp pump is used to dewater a lake. If the pump runs for 8 hours a day for 7 days a week, how  
much will it cost to run the pump for one week? Assume energy costs $0.07 per kilowatt hour.  
9. A pump station is used to lift water 50 feet above the pump station to a storage tank. The pump rate  
is 500gpm. If the pump has an efficiency of 85% and the motor has an efficiency of 90%, find each  
of the following: Water Horsepower, Brake Horsepower, Motor Horsepower, and Wire-to-Water  
Efficiency.  
10. Find the brake horsepower for a pump given the following information: Total Dynamic Head = 75  
feet, Pump Rate = 150 gpm, Pump Efficiency = 90%, Motor Efficiency = 85%  
11. Water is being pumped from a reservoir to a storage tank on a hill. The elevation difference be-  
tween water levels is 1200 feet. Find the pump size required to fill the tank at a rate of 120 gpm.  
Express your answer in horsepower.  
Solutions:  
1. Solution:  
water Hp = flow * head  
Hp  
2,200GPM 60 ft ∗  
= Water Hp = 33.3Hp  
3,960GPM ft  
pump Hp = brake Hp * pump efficiency  
33.3  
0.71  
brake Hp =  
= Brake Hp = 47Hp  
2. Solution:  
10BHp  
ηp =  
×100 = 65%  
15.4EHp  
3. Solution:  
25 Water Hp  
48 brake Hp  
ηp =  
×100 = 52%  
22.27 Dewatering math problems  
473  
4. Solution:  
Well e f ficiency = ηm ηp =0.94×0.75 = 0.705×100 = 71%  
5. Solution:  
10BHp  
= 12EHp or Input Hp  
0.85  
6. Solution:  
8.2 W Hp  
10.3 BHp  
ηp =  
×100 = 80%  
7. Solution:  
water Hp = flow * head  
Water Hp = 120 gpm1,200 ft ∗  
Hp  
= 37 Hp  
3,960 gpmft  
8. Solution:  
0.746 kW 8 hrs 7 days $0.07  
$73.1  
week  
25 Hp  
=
Hp  
day  
month kWh  
9. Solution:  
water Hp = flow * head  
Hp  
Water Hp = 500 gpm50 ft ∗  
= 6.3 WHp  
3,960 gpmft  
water Hp  
Pump efficiency =  
brake Hp  
pump Hp  
Pump efficiency  
=brake Hp =  
6.3  
0.85  
brake Hp =  
= 7.4 Hp  
brake Hp  
input Hp  
brake Hp  
motor efficiency 0.9  
7.4  
Motor efficiency =  
=input Hp =  
=
= 8.2 Hp  
474  
Chapter 22. Water Math  
Wiretowater efficiency = ηm ηp =0.9×0.85×100 = 77%  
10. Solution:  
water Hp = flow * head  
Hp  
3,960GPM ft  
150 GPM75ft∗  
= Water Hp = 2.8Hp  
pump Hp = brake Hp * pump efficiency  
2.8  
0.9  
brake Hp =  
= Brake Hp = 3.1Hp  
23. Wastewater Treatment - Future  
23.1 Water Resource Recovery Facility - WRRF  
Originally, the function of a wastewater treatment plant was collection, treatment and disposal  
which was driven solely by the need to reduce human disease and to protect the environment.  
It was soon realized that wastewater contains valuable elements like organic matter, phosphorus,  
nitrogen, rare metals and thermal energy  
The scope of wastewater treatment has now evolved to encompass recovering valuable resources  
contained in the wastewater.  
• The wastewater treatment plant is now known as a water resource recovery facility (WRRF).  
This change reflects the new focus on the products and benefits of treatment rather than its original  
and only objective - treating water for sanitation.  
The WRRF is being adapted into the concept of the circular economy in which products, mate-  
rials (and raw materials) remain in the economy for as long as possible, and waste is treated as  
secondary raw materials that can be recycled to process and re-used. This distinguishes it from a  
linear economy which is based on the: "take-make-use-dispose" system, in which waste is usually  
the last stage of the product life cycle.  
23.1.1 Water  
Municipal wastewater reuse offers the potential to significantly increase the nation’s total available  
water resources. Of the 32 billion gallons of treated wastewater discharged nationally, approxi-  
mately 12 billion gallons of treated wastewater is discharged each day to an ocean or estuary.  
WRRF is key to the use of treated wastewater, or “reclaimed” water, for beneficial purposes such  
as drinking, irrigation, or industrial uses—is one option that has helped some communities signifi-  
cantly expand their water supplies.  
Wastewater can be treated to various qualities to satisfy demand from different sectors, including  
industry and agriculture. It can be used to maintain the environmental flow, or even reused as  
 
 
 
476  
Chapter 23. Wastewater Treatment - Future  
drinking water.  
Wastewater treatment is one solution to the water scarcity issue, and also to the problem of water  
security, freeing water resources for other uses or for preservation.  
23.1.2 Energy  
In the US, municipal wastewater treatment plants consume 30 terawatt-hours per year of electricity  
which is about 0.1% of the total electrical consumption.  
WRRFs have the potential to be energy neutral or even net energy producers through comprehen-  
sive energy management approaches, incorporating efficient practices, and generating renewable  
energy from their by-products, such as biosolids.  
By making the treatment processes more energy efficient and through the recovery of chemical or  
calorific energy in the wastewater play a key role in reducing the carbon foot print of the WRRF.  
Given the amount of chemical or calorific energy contained in wastewater, the goal is for the WR-  
RFs to be energy neutral or even energy positive - which is to produce same or more energy than  
the energy needed for treatment.  
23.1.3 Nutrients  
Nutrient recovery is the practice of recovering nutrients such as nitrogen and phosphorus from used  
water streams that would otherwise be discarded/disposed and converting them into fertilizer used  
for ecological and agricultural purposes.  
Phosphate used in fertilizers is manufactured from phosphate rock using a very environmentally  
detrimental mining process. Recovery/reuse of phosphate from wastewater mitigates dependence  
on this mined mineral.  
Nutrient recovery at a WRRF is accomplished using one of following two methods:  
1. By precipitating phosphorous as struvite crystals using a dedicated reactor. The struvite crystals  
are collected and resold as fertilizer which has a resale value ranging from $100-$600 per dry ton.  
Benefits of this method include:  
• Generation of revenue from the fertilizer produced  
Reduces fouling of equipment due to precipitation of struvite formed during the solids treat-  
ment process  
• Allows for meeting NPDES nutrients discharge limits  
2. Nutrient recovery can also be achieved through the land application of biosolids. Benefits of land  
application of biosolids include:  
Provide primary nutrients - nitrogen and phosphorous and secondary nutrients such as cal-  
cium, iron, magnesium and zinc for crops.  
• The organic carbon and organic matter in the biosolids help build better soils.  
• Allows for sequestering carbon in the soil.  
23.2 Challenges to WRRF  
23.2.1 Constituents of Emerging Concern  
Constituents of Emerging Concern (CECs) include a variety of substances such as medicines,  
personal care products, flame retardants, algal toxins, micorplastics, and many others that are not  
currently federally regulated but known to occur in water.  
 
 
 
 
23.2 Challenges to WRRF  
477  
Some of these CECs, incuding hormones, PFAS, and endocrine disruptors are known to pose health  
risks to humans and aquatic life.  
Although some CECs that reach WRRFs are destroyed through wastewater treatment and solids  
processing, some recalcitrant microconstituents and their metabolites may pass through the treat-  
ment process intact and may end up in the effluent or biosolids.  
Both, the dose (concentration) of the CEC present and frequency/duration of exposure is important  
for interpreting possible risk to ecological and human health.  
The CECs move in their complex cycling through surface and groundwaters across the planet -  
from potable water to wastewater and viceversa as potable water is converted into wastewater fol-  
lowed by uptake of the treated wastewater by the potable water supply system through groundwater  
contaminated by the percolation of these CECs in land applied wastewater biosolids or through the  
CECs in the treated wastewater discharge to surface water such as a lake or river.  
The CECs concentrations in the plant influent range typically in nano-g/L to micro-g/L , in effluent  
from non-detect to nano-g/L, and in biosolids the concentrations vary from micro-g/kg to mg/kg.  
23.2.2 Decentralized and Distributed Systems  
To make the wastewater treatment systems more sustainable and to overcome the issues of a cen-  
tralized system where wastewater is collected from various areas and cities in urban areas and  
conveyed to a centrally located plant for treatment, it is imperative to consider  
Distributed systems are in different geographical locations, but are linked to a central system ei-  
ther physically, or by management. Decentralized systems can be located in a different geograph-  
ical location, but are not linked physically, or are not managed under the umbrella of a centralized  
system.  
Through the selection of correct locations and appropriate technologies, distributed or decentralized  
systems can:  
Provide environmental benefits, such as nutrient and pathogen removal  
Provide water for direct potable reuse and non-potable water in both rural and urban settings  
for purposes such as flushing, cooling and heating, landscaping, and subsurface irrigation  
drip.  
23.2.3 Climate Change  
Climate change has directly impacted water resources by altering precipitation patterns, severe  
drought and floods, snowpack amount, elevation, stream flow, and rising sea levels.  
This has created a direct need for utilities to manage local water resources to lessen the potential  
impact of climate change.  
By increasing water reuse, developing resiliency and other actions, WRRFs can be a leader in  
fighting and preparing for climate change effects.  
Driven largely by climate change factors, lower carbon footprint - the amount of carbon dioxide  
and other carbon compounds emitted due to the consumption of fossil fuels (energy), and lower  
energy demands are now being factored in when assessing wastewater treatment options.  
 
 
478  
Chapter 23. Wastewater Treatment - Future  
Source: Waste to Resource - World Bank  
24. Wastewater Treatment - Careers  
A career in wastewater treatment provides prospects of a stable and well paid employment with  
added perk of protecting the environment and the health of the community.  
Wastewater treatment plants are facing challenges to fill the positions created by the retirement of  
the baby boomers. It is estimated 30-50% of the current employees will retire within the next 10  
years.  
There are a variety of career paths which require different skill sets and training. From high school  
graduates to PhDs.  
Although actual renumeration and benefits depend on the size and location of the plant, in general  
wastewater jobs offer above average wages  
Partial List of Wastewater Career Positions  
Plant Operators  
Collections Workers  
Construction Inspectors  
Civil Engineers  
Electrical Maintenance Worker  
Laboratory Technician  
Mechanical Mntnc. Worker  
Environmental Specialists  
Electrical Engineers  
Instrumentation Mntnc. Worker  
Warehouse Workers  
Surveyors  
Mechanical Engineers  
Office Assistants  
Public Info. Specialists  
Contracts Administrators  
Health and Safety Specialist  
Planners and Schedulers  
Finance and Accounting  
Buyer  
CAD and Graphics Designers  
 
25. Glossary  
ACID : (1) A substance that tends to lose a proton. (2) A substance that dissolves in water with the forma-  
tion of hydrogen ions. (3) A substance containing hydrogen which may be replaced with metals to form  
salts. (4) A substance that is corrosive. (5) A substance that may lower pH  
ACIDITY : (A) Measure of the ability to neutralize alkaline (hi pH) substances. (B) The capacity of  
water or wastewater to neutralize bases. Acidity is expressed in milligrams per liter of equivalent calcium  
carbonate.  
ACRE FOOT : A volume of water one (1) foot deep and one (1) acre in area, or 43,560 cubic feet.  
ACTIVATED CARBON : Form of carbon processed to have small, low–volume pores that increase the  
surface area available for adsorption or chemical reactions.  
ACTIVATED SLUDGE : (A) A biological water treatment technology commonly used in municipal  
wastewater treatment systems. Sometimes private industry will harness this technique to reduce cer-  
tain pollutants, such as BOD and COD (see definitions below), but usually only due to compliance con-  
cerns. (B) Sludge withdrawn from the secondary clarifier in the activated sludge process, consisting of  
micro–organisms, non–living organic matter, and inorganic materials.  
ACTIVATED SLUDGE PROCESS : A common method of disposing of pollutants in biological wastewa-  
ters. In the process, large quantities of air are bubbled through wastewaters that contain dissolved organic  
substances in open aeration tanks. Bacteria and other types of microorganisms present in the system need  
oxygen to live, grown, and multiply in order to consume the dissolved organic "food" or pollutants in the  
waste. After several hours in a large holding tank, the water is separated from the sludge of bacteria and  
discharged from the system. Most of the activated sludge is returned to the treatment process, while the  
remainder is disposed of by one of several acceptable methods.  
ACTUATOR : Device used to operate a valve using electric, pneumatic or hydraulic means. Often used for  
remote control or sequencing of valve operations.  
ADAPTER SPOOL : An extension which is added to a short face–to–face valve, to conform to standard  
 
482  
Chapter 25. Glossary  
API 6D face–to–face dimensions.  
ADVANCED WASTE TREATMENT : (A) A treatment technology used to produce an extremely high–quality  
discharge. (B) Any process of water renovation that upgrades treated wastewater to meet specific reuse  
requirements. Typical processes include chemical treatment and pressure filtration. Also called tertiary  
treatment.  
AERATION : (A) The process of bringing about intimate contact between air and a liquid. (B) The pro-  
cess of adding air to wastewater to provide dissolved oxygen for aerobic bacterial treatment, to freshen  
wastewater and to keep solids in suspension.  
AERATION TANK : A chamber for injecting air into water.  
AEROBES : Bacteria that must have molecular (dissolved) oxygen (DO) to survive.  
AEROBIC : (A) A condition in which atmospheric or dissolved molecular oxygen is present in the aquatic  
(water) environment. (B) requiring free oxygen for respiration. Refers to types of bacteria commonly  
found in water and wastewater treatment systems.  
AEROBIC BACTERIA : Bacteria which will live and reproduce only in an environment containing oxy-  
gen which is available for their respiration (breathing), namely atmospheric oxygen or oxygen dissolved  
in water. Oxygen combined chemically, such as water molecules (H2O), cannot be used for respiration by  
aerobic bacteria.  
AIR END : A term referring to the side (or parts) of the pump that come into contact with shop/compressed  
air or natural gas. This applies to any air operated pump including Air Operated Diaphragm Pumps, air  
operated piston pumps and air operated drum pumps.  
AIR LIFT : A type of pump. This device consists of a vertical riser pipe in the wastewater or sludge to be  
pumped. Compressed air is injected into a tall piece at the bottom of the pipe. Fine air bubbles mix with  
the wastewater or sludge to form a mixture lighter than the surrounding water which causes the mixture  
to rise in the discharge pipe to the outlet. An air–lift pump works like the center of a stand in a percolator  
coffee pot.  
AIR TEST : A method of inspecting a sewer pipe for leaks. Inflatable or similar plugs are placed in the  
line, and the space between these plugs is pressurized with air. A drop in pressure indicates the line or run  
being tested has leaks.  
AIR: operated ejectors or centrifugal pumps.  
ALGAE : A class of microscopic plant life that contain chlorophyll, live floating (suspended) in water  
or are attached to rocks, walls and other surfaces, and grow and multiply through photosynthesis. Algae  
produce oxygen during sunlight hours, use oxygen during darkness and affect the pH and DO levels in  
water.  
ALGAL BLOOM : Sudden, massive growths of algae that develop in lagoons, lakes and reservoirs.  
ALIQUOT : Portion of a sample. Often an equally divided portion of a sample.  
ALKALINITY : The capacity of water to neutralize acids, a property imparted by the water’s content of  
carbonates, bicarbonates, hydroxides, and occasionally borates, silicates, and phosphates. Alkaline fluids  
have a pH value over 7.  
ALL WELDED CONSTRUCTION : Pertains to a valve construction in which the body is completely  
welded and cannot be disassembled and repaired in the field.  
483  
ANAEROBIC : A biological environment that is deficient in all forms of oxygen, especially molecular  
oxygen, nitrates and nitrites. The decomposition by microorganisms of waste organic matter in wastewater  
in the absence of dissolved oxygen is classed as anaerobic.  
ANAEROBIC BACTERIA : Bacteria that live and reproduce in an environment containing no “free” or  
dissolved oxygen. Anaerobic bacteria obtain their oxygen supply by breaking down chemical compounds  
which contain oxygen, such as sulfate (SO 2–).  
ANAEROBIC DECOMPOSITION : The decay or breaking down of organic material in an environment  
containing no “free” or dissolved oxygen.  
ANAEROBIC DIGESTION : Anaerobic bacteria (saprophytic and methane fermenters) decompose  
wastewater solids (complex organic material) in two steps into – 1) volatile acids, and 2) methane gas,  
carbon dioxide and water in the absence of dissolved oxygen. Specially designed basins, digesters, are  
used to carry out the digestion processes, prevent air from entering and to capture the methane gas. The  
sludge layer at the bottom of lagoons provides for similar solids stabilization processes.  
ANCHOR PIN : A pin welded onto the body of ball valves. This pin aligns the adapter plate and restrains  
the plate and gear operator from moving while the valve is being operated.  
ANGLE VALVE : A variation of the globe valve, in which the end connections are at right angles to each  
other, rather than being inline.  
ANIONIC FLOCCULANT : Negatively charged flocculant. Used in water treatment to aid solid / liquid  
separation  
ANOXIC : A biological environment that is deficient in molecular oxygen, but may contain chemically  
bound oxygen, such as nitrates and nitrites.  
ANOXIC : Description of an environment without oxygen. In wastewater treatment anoxic processes are  
typically used for the removal of nitrogen from wastewater.  
ANTISCALENT : Material used to control scale formation in water systems such as boiler or cooling  
water systems.  
AOD : AOD stands for Air Operated Diaphragm (pump). These types of pumps are powered by com-  
pressed air or gas, making them ideal for hazardous applications such as petroleum based products and  
other flammable materials. With certain materials of construction, such as steel or conductive plastics,  
they are easily converted into fully explosion proof (Ex–Proof) pumps. Additionally, they can pull a  
suction lift and are submersible when installed properly. AOD’s can also handle slurries with solids con-  
centrations up to 30 AODD : AODD stands for Air Operated Double Diaphragm (pump). These types  
of pumps are powered by compressed air or gas, making them ideal for hazardous applications such as  
petroleum based products and other flammable materials. With certain materials of construction, such as  
steel or conductive plastics, they are easily converted into fully explosion proof (Ex–Proof) pumps. Addi-  
tionally, they can pull a suction lift and are submersible when installed properly. AOD’s can also handle  
slurries with solids concentrations up to 30 AQUIFER : A natural underground layer of porous materials  
usually capable of yielding a supply of water.  
ARSENIC : A heavy metal commonly regulated by wastewater discharge permits, but not commonly  
found in industrial wastewaters. Other heavy metals include – Cadmium (Cd), Chromium(Cr), Copper  
(Cu), Lead (Pb), Nickel (Ni), and Zinc (Zn).  
ASPHYXIATION : An extreme condition often resulting in death due to a lack of oxygen and excess  
484  
Chapter 25. Glossary  
carbon dioxide in the blood from any cause.  
AVAILABLE CHLORINE : The amount of chlorine available in compound chlorine sources compared  
with that of elemental (liquid or gaseous) chlorine.  
AVERAGE MONTHLY DISCHARGE LIMITATION : The highest allowable discharge over a calendar  
month  
AVERAGE WEEKLY DISCHARGE LIMITATION : The highest allowable discharge over a calendar  
week.  
B.R.V. : BODY RELIEF VALVE – A relief valve (optional) installed on ball valves used in liquid service  
to provide for the relief of excess body pressure caused by thermal expansion.  
BACK WASH : Part of water filter, ion exchange or softener cycle that lifts up media bed to release and  
wash away dirt and other foulants.  
BACKFILL : (1) Material used to full in a trench or excavation. (2) The act of filling a trench or excava-  
tion, usually after a pipe or some type of structure has been placed in the trench or excavation.  
BACKFILL COMPACTION : (1) Tamping, rolling or otherwise mechanically compressing material used  
as backfill for a trench of excavation. Backfill is compressed to increase its density so that it will support  
the weight of machinery or other loads after the material is in place excavation. (2) Compaction of a  
backfill material can be expressed as a percentage of the maximum compatibility, density or load capacity  
of the material being used.  
BACKFLOW : A reverse flow condition, created by a difference in water pressures, which causes water  
to flow back into the distribution pipes of a potable water supply from any source or sources other than an  
intended source. Also see BACKSIPHONAGE.  
BACKFLUSHING : A procedure used to wash settled waste matter off upstream to prevent odors from  
developing after a main line stoppage has been cleared.  
BACKSEAT : A Shoulder on the stem of a valve which seals against a mating surface inside the bonnet to  
permit replacement, under pressure, of stem seals or packing.  
BACKSIPHONAGE : A form of backflow caused by a negative or below atmospheric pressure within a  
water system. Also see BACKFLOW.  
BACTERIA : Bacteria are microscopic living organisms They are a group of universally distributed, rigid,  
essentially unicellular, microscopic organisms lacking chlorophyll. They are characterized as spheroids,  
rod–like, or curved entities, but occasionally appearing as sheets, chains, or branched filaments.  
BAFFLE : A flat board or plate, deflector, guide or similar device constructed or placed in flowing water,  
wastewater, or slurry systems to cause more uniform flow velocities, to absorb energy, and to divert, guide,  
or agitate liquids (water, chemical solutions, slurry).  
BALL : The spherical closure element of a ball valve.  
BALL CHECK : A fitting with a small ball that seals against a seat preventing flow in one direction and  
allowing flow in the other direction.  
BALL VALVE : A valve using a spherical closure element (ball) which is rotated thru 90° to open and  
close the valve.  
BALLING : A method of hydraulically cleaning a sewer or storm drain by using the pressure of a water  
head to create a high cleansing velocity of water around the ball. In normal operation, the ball is restrained  
485  
by a cable while water washes past the ball at high velocity. Special sewer cleaning balls have an outside  
tread that causes them to spin or rotate, resulting in a “scrubbing” action of the flowing water along the  
pipe wall.  
BAR RACK : A screen composed of parallel bars, either vertical or inclined, placed in a sewer or other  
waterway to catch debris. The screenings may be raked from it.  
BARREL : (1) The cylindrical part of a pipe that may have a bell on one end. (2) The cylindrical part of a  
manhole between the cone at the top and the shelf at the bottom.  
BASE : (1) A substance which takes up or accepts protons. (2) A substance which dissociates (separates)  
in aqueous solution to yield hydroxyl ions (OH–). (3) A substance containing hydroxyl ions which reacts  
with an acid to form a salt or which may react with metalsvto form precipitates. (4) A substance that may  
raise pH.  
BASE : A substance that takes up or accepts protons, dissociates in water to produce hydroxyl (OH–) ions,  
reacts with metals and is corrosive.  
BDV : BLOW DOWN VALVE – A small ball valve that is installed on the aboveground end of an ex-  
tended drain line. This valve also serves to vent body cavity pressure in the "block and bleed" mode.  
BEDDING : The prepared base or bottom of a trench or excavation on which a pipe or other underground  
structure is supported.  
BEDDING COMPACTION : (1) Tamping, rolling or otherwise mechanically compressing material used  
as bedding for a pipe or other underground structure to a density that will support expected loads. (2)  
Bedding compaction can be expressed as a percentage of the maximum load capacity of the bedding  
material. (3) Bedding compaction also can be expressed in load capacity or pounds per square foot.  
BEDDING GRADE : (1) In a gravity–flow sewer system, pipe bedding is constructed and compacted to  
the design grade of the pipe. This is usually expressed in a percentage. A 0.5 percent grade would be a  
drop of one–half of foot per hundred feet of pipe. (2) Bedding grade for a gravity–flow sewer pipe can  
also be specified as elevation above mean sea level at specific points.  
BELL : (1) In pipe fitting, the enlarged female end of a pipe into which the male end fits. (2) In plumbing,  
the expanded female end of a wiped joint.  
BELL: AND–SPIGOT JOINT – A form of joint used on pipes which have an enlarged diameter or bell  
at one end, and a spigot at the other which fits into and is laid in the bell. The joint is then made tight by  
lead, cement, rubber O–ring, or other jointing compounds or materials.  
BELLEVILLE SPRING : A spring resembling a dished washer, used in some ball valves to push the seats  
against the ball.  
BERM : The earthen dike that surrounds ponds, lagoons and containment areas for hazardous material.  
BEST EFFICIENCY POINT (B.E.P.) : The point on a pump’s performance curve that corresponds to the  
highest efficiency.  
BEVEL GEAR OPERATOR : Device facilitating operation of a gate or globe valve by means of a set of  
bevel gears having the axis of the pinion gear at right angles to that of the larger ring gear. The reduction  
ratio of this gearsetdetermines the multiplication of torque achieved.  
BHP : BHP is the actual amount of horsepower being consumed by the pump as measured on a pony  
brake or dynamometer.  
486  
Chapter 25. Glossary  
BIOCHEMICAL OXYGEN DEMAND (BOD) : A quantitative measure of the oxygen needed by bacteria  
and micro–organisms for the biological oxidation of organic wastes in a unit volume of wastewater. BOD  
is generally measured in milligrams per liter (mg/l) of oxygen consumed over a five day period. Although  
complete biological decomposition of organic waste requires about 20 days, the five day BOD is about  
two–thirds of the total oxygen required and, therefore, is a practical measure of waste concentration. In  
waste treatment language, BOD is most frequently stated as the percentage removed during treatment, or  
remaining after treatment.  
BIOCIDE : Chemical substance designed for killing living organisms in water. Often characterized by  
type of organism killed – bactericide, fungicide or algaecide.  
BIOLOGICAL OXIDATION : The process by which bacteria and other types of microorganisms consume  
dissolved oxygen and organic substances in biological wasterwater. The energy released is then used to  
convert organic carbon into carbon dioxide and cellular material.  
BIOMASS : Amass or clump of living organisms feeding on wastes in wastewater, dead organisms and  
other debris. The mass may protect the organisms, as well as store food supplies. Also called ZOOGLEAL  
MASS.  
BIOSOLIDS : A primarily organic solid product, produced by wastewater treatment processes, that can be  
beneficially recycled. The word biosolids is replacing the word sludge.  
BIOSOLIDS CAKE : Solid discharge from a dewatering apparatus.  
BIT : (1) Cutting blade used in rodding (pipe cleaning) operations. (2) Cutting teeth on the auger head of a  
sewer boring tool.  
BLANK : A bottle containing only dilution water or distilled water, but the sample being tested is not  
added. Tests are frequently run on a SAMPLE and a BLANK and the differences are compared.  
BLOCK AND BLEED : The capability of obtaining a seal across the upstream and downstream seat rings  
of a valve when the body pressure is bled off to atmosphere thru blow down valves or vent plugs. Useful  
in testingfor integrityof seat seals and in accomplishing minor repairs under pressure.  
BLOCKAGE : (1) Partial or complete interruption of flow as a result of some obstruction in a sewer. (2)  
When a collection system becomes plugged and the flow backs up, “blockage.”  
BLOW DOWN (BLEED: off) – terms to describe the deliberate rejection of water from a system such  
as boiler or cooling water system. Typically done to control system’s water total dissolved system or  
conductivity.  
BLUE: GREEN ALGAE – Varieties of algae characterized by their bluish–green color. The appearance of  
blue–green algae indicates unhealthy conditions in lagoon cells, often associated with organic overloading  
and lack of adequate dissolved oxygen.  
BOD : Biochemical Oxygen Demand. The rate at which organisms use the oxygen in water or wastewater  
while stabilizing decomposable organic matter under aerobic conditions. In decomposition, organic matter  
serves as food for the bacteria and energy results from its oxidation. BOD measurements are used as a  
measure of the organic strength of wastes in water.  
BODY : The principal pressure containing part of a valve, in which the closure element and seats are  
located.  
BOLTED BONNET : A bonnet which is connected to a valve body with bolts or studs and nuts.  
487  
BOLTED CONSTRUCTION : Describes a valve construction in which the pressure shell elements are  
bolted together, and thus can be taken apart and repaired in the field.  
BONNET : The top part of a valve, attached to the body, which contains the packing gland, guides the  
stem, and adapts to extensions or operators.  
BORE (OR PORT) : The inside diameter of the smallest opening through a valve, e.g., inside diameter of  
a seat ring, diameter of hole through ball in a ball valve.  
BRANCH MANHOLE : A sewer or drain manhole which has more than one pipe feeding into it. A  
standard manhole will have one outlet and one inlet. A branch manhole will have one outlet and two or  
more inlets.  
BRANCH SEWER : A sewer that receives wastewater from a relatively small area and discharges into a  
main sewer servicing more than one branch sewer area.  
BUBBLE: TIGHT SHUT–OFF – A phrase used in describing the sealing ability of a valve. During air  
pressure testing of a new valve in the closed position, leakage past the seats is collected and bubbled thru  
water. To qualify as "bubble tight," no bubbles should be observed in a prescribed time span.  
BUCKET : (1) A special device designed to be pulled along a sewer for the removal of debris from the  
sewer. The bucket has one end open with the opposite end having a set of jaws. When pulled from the jaw  
end, the jaws are automatically opened. When pulled from the other end, the jaws close. In operation, the  
bucket is pulled into the debris from the jaw end and to a point where some of the debris has been forced  
into the bucket. The bucket is then pulled out of the sewer from the other end, causing the jaws to close  
and retain the debris. Once removed from the manhole, the bucket is emptied and the process repeated.  
(2) A conventional pail or bucket used in BUCKETING OUT and also for lowering and raising tools and  
materials from manholes and excavations.  
BUCKET BAIL : The pulling handle on a bucket machine.  
BUCKET MACHINE : A powered winch machine designed for operation over a manhole. The machine  
controls the travel of buckets used to clean sewers.  
BUCKETING OUT : An expression used to describe removal of debris from a manhole with a pail on  
a rope. In balling or high–velocity cleaning of sewers, debris is washed into the downstream manhole.  
Removal of this debris by scooping it into pails and hauling debris out is called “bucketing out.”  
BUFFER : A solution or liquid whose chemical makeup neutralizes acids or bases without a great change  
in pH.  
BULKING : Clouds of billowing sludge that occur throughout secondary clarifiers and sludge thickeners  
when the sludge does not settle properly. In the activated sludge process bulking is usually caused by  
filamentous bacteria or bound water.  
BULKING SLUDGE : A phenomenon that occurs in activated sludge plants whereby the sludge occupies  
excessive volumes and will not concentrate readily. This condition refers to a decrease in the ability of  
the sludge to settle and consequent loss over the settling tank weir. Bulking in activated sludge aeration  
tanks is caused mainly by excess suspended solids (SS) content. Sludge bulking in the final settling tank  
of an activated sludge plant may be caused by improper balance of the BOD load, SS concentration in the  
mixed liquor, or the amount of air used in aeration.  
BURIED SERVICE : An application in which valves are installed in lines which are buried below ground  
level.  
488  
Chapter 25. Glossary  
BUTT WELD END (BWE) : The end connection of a valve suitably prepared for butt welding to a con-  
necting pipe.  
BUTTERFLY VALVE : A short face–to–face valve which has a movable vane, in the center of the flow  
stream, which rotates 90 degrees as the butterfly valve opens and closes.  
BVR : BALL VALVE REGULATOR – An automatic throttling valve controlling flow or pressure in a  
pipeline; comprising a package involving al ball valve actuator, positioner, and controlling instrument.  
BYPASS : A system of pipes and valves permitting the diversion of flow or pressure around a line valve.  
BYPASS : A pipe, valve, gate, weir, trench or other device designed to permit all or part of a wastewater  
flow to be diverted from usual channels or flow. Sometimes refers to a special line which carries the flow  
around a facility or device that needs maintenance or repair.  
BYPASSING : The act of causing all or part of a flow to be diverted from its usual channels. In a wastewa-  
ter treatment plant, overload flows should be bypassed into a holding pond for future treatment. 224  
CADMIUM : A heavy metal commonly regulated by wastewater discharge permits and typically found  
in the metal finishing industry. Other heavy metals include – Arsenic (As), Chromium(Cr), Copper (Cu),  
Lead (Pb), Nickel (Ni), and Zinc (Zn).  
CAKE SOLID DISCHARGE RATE : The dry solids cake discharge from a centrifuge, which is expressed  
as: dry cake solids discharge rate = (dry solids feed rate) x (solids recovery).  
CALCIUM AND MAGNESIUM SOAPS, MINERAL OILS, AND CERTAIN OTHER NON: fatty mate-  
rial which tend to separate from water and coagulate as floatables or scums.  
CARBON DIOXIDE : A common gas, CO2, found abundantly in air, is a product of bacterial respiration  
and used by algae in photosynthesis. The concentration of carbon dioxide in the lagoon water governs the  
pH of the lagoon.  
CARCINOGEN : Any substance that tends to produce cancer in an organism.  
CASING : The body of the pump which encloses the impeller. Primarily used in reference to centrifugal  
pumps.  
CAST : The form of a particular part of a valve, where the basic shape is formed by molding rather than  
fabricating.  
CASTING : A product or the act of producing a product made by pouring molten metal into a mold and  
allowing it to solidify, thus taking the shape of the mold.  
CATCH BASIN : A chamber or well used with storm or combined sewers as a means of removing grit  
which might otherwise enter and be deposited in sewers.  
CATIONIC FLOCCULANT : Positively charged high molecular weight polyelectrolyte water soluble  
organic polymer designed to agglomerate solids in water substrates.  
CAVITATION : The formation and collapse of a gas pocket or bubble on the blade of an impeller or gate  
of a valve. The collapse of the bubble drives water into the impeller or gate with a terrific force that can  
cause pitting of the surface. Cavitation is indicated by loud hammering noises.  
CENTRIFUGAL FORCE : A force associated with a rotating body. In the case of a pump, the rotating  
impeller pushes fluid on the back of the impeller blade, imparting motion. Since the motion is circular  
there is a centrifugal force associated with it. The force pushes the fluid against a fixed pump casing  
thereby pressurizing the fluid and forcing it through the outlet.  
489  
CENTRIFUGAL PUMP : Centrifugal pumps are the most common type of pump in use today throughout  
the world. A centrifugal pump is a rotodynamic pump that uses a rotating impeller to increase the velocity  
of a fluid. Centrifugal pumps are commonly used to move liquids through a piping system. The fluid  
enters the pump impeller along or near to the rotating axis and is accelerated by the impeller, flowing  
radially outward into a diffuser or volute chamber, from there it exits into the downstream piping system.  
A centrifugal pump works by the conversion of the rotational kinetic energy, typically from an electric  
motor or engine, to an increased static fluid pressure. This action is described by Bernoulli’s principle.  
The rotation of the pump impeller imparts kinetic energy to the fluid as it is drawn in from the impeller  
eye (center) and is forced outward through the impeller vanes to the periphery. As the fluid exits the  
impeller, the fluid kinetic energy (velocity) is then converted to (static) pressure due to the change in area  
the fluid experiences in the volute section. Typically, the volute shape of the pump casing (increasing  
in volume), or the diffuser vanes (which serve to slow the fluid, converting to kinetic energy in to flow)  
are responsible for the energy conversion. The energy conversion results in an increased pressure on the  
downstream side of the pump, causing flow.  
CENTRIFUGE : A mechanical device that uses centrifugal or rotational forces to separate solids from  
liquids.  
CHAIN WHEEL OPERATED VALVE : An overhead valve operated by a chain drive wheel instead of a  
handwheel.  
CHARACTERIZED GATE OR BALL : A ball or gate, the shape of whose port has been specially altered  
to provide a specific throttling capability.  
CHECK VALVE : A one–directional valve which is opened by the fluid flow in one direction and closed  
automatically when the flow stops or is reversed.  
CHEMICAL GROUT : Two chemical solutions that form a solid when combined. Solidification time is  
controlled by the strength of the mixtures used and the temperature.  
CHEMICAL OXYGEN DEMAND (COD) : A quantitative measure of the amount of oxygen required to  
oxidize all organic compounds in a unit volume on wastewater – non–biodegradable as well as the BOD.  
The COD level can be determined more readily than BOD, but this measurement does not indicate how  
much of the waste can be decomposed by biological oxidation.  
CHLORINATION : The application of chlorine to water, sewage, or industrial wastes, generally for  
the purpose of disinfection, but frequently for accomplishing other chemical or biological wasterwater  
treatment results.  
CHLORINATOR : A metering device which is used to add chlorine to water.  
CHLORINE CONTACT UNIT : A baffled basin that provides sufficient time for disinfection to occur.  
CHLORINE DEMAND : Chlorine demand is the difference between the amount of chorine added to  
wastewater and the amount of residual chlorine remaining after a given contact time. Chlorine demand  
may change with dosage, time temperature, pH, and nature and amount of the impurities in the water.  
CHLORINE REQUIREMENT : The amount of chlorine which is needed for a particular purpose. Some  
reasons for adding chlorine are reducing the number of coliform bacteria (Most Probable Number), obtain-  
ing a particular chlorine residual, or oxidizing some substance in the water. In each case a definite dosage  
of chlorine will be necessary. This dosage is the chlorine requirement.  
CHLORINE RESIDUAL : The amount of free chlorine remaining after meeting chlorine demand under  
490  
Chapter 25. Glossary  
given conditions and is necessary to complete disinfection.  
CHOPPER PUMP : A chopper pump is a centrifugal pump, which is equipped with a cutting system  
to facilitate chopping/maceration of solids that are present in the pumped liquid. The main advantage  
of this type of pump is that it prevents clogging of the pump itself and of the adjacent piping, as all the  
solids and stringy materials are macerated by the chopping system. Chopper pumps exist in various  
configurations, including submersible and dry–installed design and they are typically equipped with an  
electric motor to run the impeller and to provide torque for the chopping system. Due to its high solids  
handling capabilities, the chopper pump is often used for pumping sewage, sludge, manure slurries, and  
other liquids that contain large or tough solids.  
CHROMIUM : A heavy metal commonly regulated by wastewater discharge permits and found in met-  
als–related industries and products (including stainless steel). It is typically regulated in two forms – total  
chromium and hexavalent chromium. Other heavy metals include – Arsenic (As), Cadmium (Cd), Copper  
(Cu), Lead (Pb), Nickel (Ni), and Zinc (Zn).  
CITY GATE : CITY GATE STATION – The metering and pressure reducing station where gas is trans-  
ferred from a high pressure cross–country transmission line to a low pressure distribution piping system  
within a city.  
CLAPPER : The hinged closure element of a swing check valve.  
CLARIFICATION : Any process or processes used to reduce the concentration of suspended matter in a  
liquid, such as quiescent settling or sedimentation. Lagoons provide clarification across the cells and in  
quiescent zones in aerated systems, allowing solids to settle into a sludge layer  
CLARIFIER : A large, circular or rectangular tank that separates solids from the waste stream by settling  
or flotation.  
CLEAN IN PLACE (CIP) : Method of cleaning the interior surfaces of pipes, vessels, process equipment,  
filters and associated fittings, without disassembly.  
CLEAN WATER ACT : Federal legislation passed in 1972 creating the Environmental Protection Agency,  
requiring a nationwide system for controlling pollutant discharges and providing for construction and  
regulation of publicly owned treatment works.  
CLEANOUT : An opening (usually covered or capped) in a wastewater collection system used for insert-  
ing tools, rods or snakes while cleaning a pipeline or clearing a stoppage.  
CLOSURE ELEMENT : The moving part of a valve, positioned in the flowstream which controls flow  
thru the valve. Ball. Gate, Plug, Clapper, Disc, etc., are specific names for closure elements.  
CO: precipitation – Term to describe compound used in water treatment to aid precipitation of substances  
normally soluble under the conditions employed. Common co–precipitants used in water are iron, alu-  
minum, calcium and magnesium.  
COAGULANTS : Chemicals that cause very fine particles to clump (floc) together into larger particles.  
This makes it easier to separate the solids from the water by settling, skimming, draining or filtering.  
COAGULATION : The agglomeration of colloidal or suspended matter brought about by the addition of  
some chemical to the liquid, by contact, or by other means. Involves destabilization for repulsive electrical  
charges to permit agglomeration of colloid particles in water. This process aids the clarification of water.  
COLIFORM : A type of bacteria. The presence of coliform–group bacteria is an indication of possible  
pathogenic bacterial contamination. The human intestinal tract is one of the main habitats of coliform bac-  
491  
teria. They may also be found in the intestinal tracts of warm–blooded animals, and in plants, soil, air, and  
the aquatic environment. Fecal coliforms are those coliforms found in the feces of various warm–blooded  
animals; whereas the term “coliform” also includes various other environmental sources.  
COLIFORM : A type of bacteria. The presence of coliform–group bacteria is an indication of possible  
pathogenic bacterial contamination. The human intestinal tract is one of the main habitats of coliform bac-  
teria. They may also be found in the intestinal tracts of warm–blooded animals, and in plants, soil, air, and  
the aquatic environment. Fecal coliforms are those coliforms found in the feces of various warm–blooded  
animals; whereas the term “coliform” also includes various other environmental sources.  
COLIFORM BACTERIA : Live in everyone’s intestinal track. They are considered non–pathogenic.  
COLIFORM ORGANISMS : A group of bacteria recognized as indicators of fecal pollution (see also  
escherichia coliform).  
COLLECTION SYSTEM : A network of pipes, manholes, cleanouts, traps, siphons, lift stations and  
other structures used to collect all wastewater and wastewater–carried wastes of an area and transport  
them to a treatment plant or disposal system. The collection system includes land, wastewater lines and  
appurtenances, pumping stations and general property.  
COLORIMETRIC MEASUREMENT : A means of measuring unknown chemical concentrations in water  
by MEASURING A SAMPLE’S COLOR INTENSITY. The specific color of the sample, developed by  
addition of chemical reagents, is measured with a photoelectric colorimeter or is compared with “color  
standards” using, or corresponding with, known concentrations of the chemical.  
COLORIMETRIC MEASUREMENT : A means of measuring unknown chemical concentrations in water  
by MEASURING A SAMPLE’S COLOR INTENSITY. The specific color of the sample, developed by  
addition of chemical reagents, is measured with a photoelectric colorimeter or is compared with “color  
standards” using, or corresponding with, known concentrations of the chemical.  
COMBINED SEWER : Carries both sanitary sewage and storm water run–off.  
COMMINUTOR : A device used to reduce the size of the solid chunks in wastewater by shredding (com-  
minuting). The shredding action is like many scissors cutting or chopping to shreds all the large influent  
solids material in the wastewater.  
COMMUNITY WASTEWATER SYSTEM : A public wastewater system which has at least 15 service  
connection or treats 5,000 gallons or more of wastewater per day. The term “community wastewater sys-  
tem” is used only to identify the public wastewater systems which must be operated by certified operators.  
COMPOSITE (PROPORTIONAL) SAMPLE : A collection of individual samples obtained at regular  
intervals during a 24–hour period. Each individual sample is combined with the others in proportion to the  
rate of flow when the sample was collected. The resulting mixture, or composite, forms a representative  
sample and is analyzed to determine the average conditions during the sampling period.  
COMPUTED PER CAPITA CONTRIBUTION : The computed wastewater contribution from a domestic  
area, based on the population of the area. In the United States, the daily average wastewater contribution is  
considered to be 100 gallons per capita per day (100GPCD).  
COMPUTED TOTAL CONTRIBUTION : The total anticipated load on a wastewater treatment plant or  
the total anticipated flow in any collection system area based on the combined computed contributions of  
all connections to the system.  
CONCENTRATE : The high TDS discharge from a reverse osmosis filtration process.  
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Chapter 25. Glossary  
CONCRETE CRADLE : A device made of concrete that is designed to support sewer pipe. 225  
CONDENSATE : steam that has lost heat and condensed into water.  
CONDUCTIVITY : Transmittance of an electric current through water. Usually measured in microsiemens  
per centimeter (uS/cm) or micromho per centimeter (umho/cm).  
CONFINED SPACE : Confined space means a space that: A. Is large enough and so configured that an  
employee can bodily enter and perform assigned work; and B. Has limited or restricted means for entry or  
exit; and C. Is not designed for continuous employee occupancy.  
CONTAMINATION : The introduction into water of microorganisms, chemicals, toxic substances, wastes,  
or wastewater in a concentration that makes the water unfit for its next intended use.  
CONTROL VALVE : A valve that controls a process variable, such as pressure, flow or temperature by  
modulating its opening in response to a signal from a controller.  
CONTROL VOLUME : Limits imposed for the theoretical study of a system. The limits are usually set to  
intersect the system at locations where conditions are known.  
CONTROLLER : A device that measures a controlled variable, compares it with a predetermined setting  
and signals the actuator to read just the opening of the valve in order to re–establish the original control  
setting.  
COPPER : A heavy metal commonly regulated by wastewater discharge permits. It is found in the  
metal finishing and electrical industries. Other heavy metals include – Arsenic (As), Cadmium (Cd),  
Chromium(Cr), Lead (Pb), Nickel (Ni), and Zinc (Zn).  
CORROSION : The gradual decomposition or destruction of a material due to chemical action, often due  
to an electrochemical reaction. Corrosion starts at the surface of a material and moves inward, such as the  
chemical action upon manholes and sewer pipe materials.  
CORROSION INHIBITOR : chemical additive designed to control / minimize metal corrosion in water  
system.  
COULISSE : Of or using runners or slides as a guiding mechanism; as in a "Coulisse" style gate valve.  
COUPLING : (1) A threaded sleeve used to connect two pipes. (2) A device used to connect two adjacent  
parts, such as pipe coupling, hose coupling or drive coupling.  
COUPON : A steel specimen inserted into wastewater to measure the corrosiveness of the wastewater.  
The rate of corrosion is measured as the loss of weight of the coupon or change in its physical charac-  
teristics. Measure the weight loss (in milligrams) per surface area (in square decimeters) exposed to the  
wastewater per day.  
CREST : The bottom edge of a weir plate.  
CROSS CONNECTION : A connection between a drinking (potable) water system and an unapproved  
water supply. For example, if you have a pump moving nonpotable water and hook into the drinking water  
system to supply water for the pump seal, a cross connection or mixing between the two water systems  
can occur. This mixing may lead to contamination of the drinking water.  
CROSS CONNECTION : A connection between a drinking (potable) water system and an unapproved  
water supply. For example, if you have a pump moving nonpotable water and hook into the drinking water  
system to supply water for the pump seal, a cross connection or mixing between the two water systems  
can occur. This mixing may lead to contamination of the drinking water.  
493  
CROSS CONNECTION : A connection between a storm drain system and a sanitary collection system.  
CROSS: CONNECTION – A connection between a drinking water system and an unapproved system.  
CRUSTACEANS : A class of microscopic water animals that consume large quantities of bacteria and  
algae.  
CRYOGENIC VALVE : A valve capable of functioning at cryogenic temperatures.  
CYANIDE : A toxic element often found in wastewater from metal finishing industries. It’s commonly  
regulated by wastewater permits.  
CYCLE : A single complete operation or process returning to the starting point. A valve, stroked from full  
open to full close and back to full open, has undergone one cycle.  
CYCLES OF CONCENTRATION : Ratio of boiler or cooling water to make up (feed water). Typically  
measured by monitoring total dissolved solids, conductivity, silica or chlorides.  
CYLINDER OPERATOR : A power–piston valve operator using either hydraulic or pneumatic pressure.  
A sealed piston converts applied pressure into a linear piston rod (stem) motion.  
DAILY DISHARGE : The discharge of a pollutant measured during a calendar day or any 24 – hour  
period that reasonably represents a calendar day for the purposes of sampling.  
DAILY MAXIUM DISCHARGE : The highest allowable value for a daily discharge.  
DAPHNIA : A crustacean commonly found in wastewater lagoons.  
DATUM PLANE : A reference plane. A conveniently accessible known surface from which all vertical  
measurements are taken or referred to.  
DEADEND MANHOLE : A manhole located at the upstream end of a sewer and having no inlet pipe.  
DEBRIS : Any material in wastewater found floating, suspended, settled, or moving along the bottom of  
a sewer. This material may cause stoppages by getting hung up on roots or settling out in a sewer. Debris  
includes grit, paper, rubber, silt, and all materials except liquid.  
DECHLORINATION : The removal of chlorine from the effluent of a treatment plant.  
DECHLORINATION : The removal of chlorine from the effluent of a treatment plant.  
DECHLORINATION : The removal of chlorine from the effluent of a treatment plant.  
DEMULSIFIER (EMULSION BREAKER) : Chemical additive that destroys the emulsifying characteris-  
tics water. Typically used separate stabilized (emulsified) oil in water.  
DENITRIFICATION : A biological process by which nitrate is converted to nitrogen gas.  
DENITRIFICATION : (1) The anoxic biological reduction of nitrate nitrogen to nitrogen gas. (2) The  
removal of some nitrogen from a system. (3) An anoxic process that occurs when nitrite or nitrate ions are  
reduced to nitrogen gas and nitrogen bubbles are formed as a result of this process.  
DENITRIFICATION : (1) The anoxic biological reduction of nitrate nitrogen to nitrogen gas. (2) The  
removal of some nitrogen from a system. (3) An anoxic process that occurs when nitrite or nitrate ions are  
reduced to nitrogen gas and nitrogen bubbles are formed as a result of this process.  
DENITRIFICATION : An anaerobic process that occurs when nitrite and nitrate ions are reduced to  
nitrogen gas and bubbles are formed. These bubbles attach to sludge flocs, causing rising sludge that floats  
to the surface of secondary clarifiers.  
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Chapter 25. Glossary  
DENITRIFICATION : Wastewater treatment process involved in the biological removal of nitrogen in  
which the nitrite (NO2) is converted to nitrogen gas (N2)  
DEQ : Department of Environmental Quality.  
DETENTION TIME : The theoretical time that water may stay in a basin such as lagoon. It is the total  
volume of the lagoon divided by the flow rate. Usually expressed in days of time or in hours. The theoreti-  
cal time water remains in a tank at a given discharge. The time required to fill a tank at a given flow or the  
theoretical time required for a given flow of wastewater to pass through a tank.  
DETRITUS : The heavy, coarse mixture of grit and organic material carried by wastewater. (also called  
grit).  
DEWATER : To drain or remove water from an enclosure. A structure may be dewatered so that it can be  
inspected or repaired. Dewater also means draining or removing water from sludge to increase the solids  
concentration. 226  
DEWATERING : Removing free water from a sludge or slurry to form a high solids cake. belt filter  
presses, centrifuges, rotary fan presses and vacuum presses are dewatering devices.  
DEWATERING PUMP : Any pump capable of removing water from an unwanted area. They are usually  
small, portable pumps that run on single phase power, compressed air or a small engine, but can be large  
permanently installed units as well.  
DIAPHRAGM : A round, thin flexible sealing device secured and sealed around its outer edge – and  
sometimes around a central hole in the diaphragm – with its unsupported area free to move by flexing.  
DICHARGE MONITORING REPORT (DMR) : The monthly report required by the treatment plant’s  
NPDES / OPDES discharge permit.  
DIFFUSED: AIR AERATION – A diffused air activated sludge plant takes air, compresses it, and then  
discharges the air below the water surface of the aerator through some type of air diffusion device.  
DIFFUSER : A device used to break the air stream from the blower system into fine bubbles in an aeration  
tank or reactor.  
DIGESTER : A tank in which sludge is placed to allow decomposition by microorganisms. Digestion may  
occur under anaerobic (more common) or aerobic conditions.  
DIGESTION : The biochemical decomposition of organic matter that results in the formation of mineral  
and simpler organic compounds.  
DIP : A point in the sewer pipe where a drain grade defect results in a puddle of standing water when  
there is no flow.  
DIP TUBE : Extending the blow down valve on large gate valves requires a tube which is located inside  
of the valve. The tube is called the "dip tube" and extends through the bonnet to the bottom of the body  
cavity.  
DISC : The closure element of a globe angle or small regulator valve. The disc (sometimes referred to  
as "valve," "poppet" or "plug") moves to and from the seat in a direction perpendicular to the seat face.  
Depends on stem force for tight shutoff.  
DISCHARGE STATIC HEAD : The difference in elevation between the liquid level of the discharge tank  
and the centerline of the pump. This head also includes any additional head that may be present at the  
discharge tank fluid surface.  
495  
DISINFECTION : The process designed to kill most microorganisms in water, including the destruction  
or inactivation of pathogenic bacteria. Disinfection differs from sterilization which destroys all living  
forms.  
DISINFECTION : The process designed to kill or inactivate most microorganisms in wastewater, in-  
cluding essentially all pathogenic (disease–casing) bacteria. There are several ways to disinfect, with  
chlorination being the most frequently used in water and wastewater treatment plants.  
DISINFECTION : The process designed to kill or inactivate most microorganisms in wastewater, in-  
cluding essentially all pathogenic (disease–casing) bacteria. There are several ways to disinfect, with  
chlorination being the most frequently used in water and wastewater treatment plants.  
DISINFECTION : The process designed to kill or inactivate most microorganisms in water, including  
essentially all pathogenic (disease–causing) bacteria. There are several ways to disinfect, with chlorine  
being the most frequently used method in both water and wastewater systems.  
DISPERSANT : A non–surface active compound or an active substance added to a suspension, usually a  
mix, to increase the separation of particles and to prevent subsiding or clumping.  
DISSOLVED AIR FLOTATION : A physical/chemical wastewater treatment technology that can be  
cost–effectively used by industry to remove FOG (fats, oils and grease), suspended solids, and some  
metals.  
DISSOLVED AIR FLOTATION CLARIFIER (DAF) : a piece of equipment that used dissolved air to  
float suspended solids from water. Typically used when the suspended solids have a lower density than  
water.  
DISSOLVED OXYGEN (DO) : The oxygen dissolved in water, wastewater, or other liquid. DO is mea-  
sured in milligrams per liter. If the DO of a sample of water is 2 mg/L, it means that there are 2lbs of  
oxygen in 1 mil lb of water.  
DISSOLVED SOLIDS : The salts and other residues left after evaporation of water that has been passed  
through a laboratory filter. Dissolved solids cannot be filtered out. Some colloidal solids may not be in  
true solution, but if they pass through the standard membrane filter, they are considered dissolved solids.  
(See suspended solids)  
DIURNAL : Having a daily cycle; usually a 24–hour period from 12:00am to 12:00pm.  
DO : Dissolved Oxygen – An indication of how much oxygen is present in water. If a facility discharges  
directly to a stream or river, it will usually have a permit limit related to dissolved oxygen.  
DRAGLINE : A machine that drags a bucket down the intended line of a trench to dig or excavate the  
trench. Also used to dig holes and move soil or aggregate.  
DRAIN PLUG : A fitting at the bottom of a valve, the removal of which permits draining and flushing the  
body cavity.  
DREDGE PUMP : A dredge pump is a submersible, centrifugal pump capable of handling high solids  
concentrations and is typically used for clearing out and/or deepening harbors and waterways. The mate-  
rial being moved (i.e.) sand, dirt, soil, etc.) is carried away along with the water it is suspended in.  
DRIVE PINS : The two pins which fit into the bottom of a ball valve stem and engage corresponding  
holes in the ball. As the operator turns the stem, the drive pins turn the ball.  
DROOP : A drop in set (outlet) pressure of a regulator or control valve due to the travel of its valve or pop-  
496  
Chapter 25. Glossary  
pet, as the required flow increases from low to maximum. A slight change in the control spring length due  
to the valve travel, will result in spring force variations, translating into a change of set (outlet) pressure.  
DROP MANHOLE : A main line or house service line lateral entering a manhole at a higher elevation  
than the main flow line or channel. If the higher elevation flow is routed to the main manhole channel  
outside of the manhole, it is called an “outside drop.” If the flow is routed down through the manhole  
barrel, the pipe down to the manhole channel is called an “inside drop.”  
DRY WELL : A dry room or compartment in a lift station, near or below the water level, where the pumps  
are located.  
DUCKWEED : A water plant with single small leaf that floats and accumulates on the surface of lagoons.  
E. COLI OR ESCHERICHIA COLIFORM : A species of bacteria found in large numbers in the intestinal  
tract of warm–blooded animals.  
EASEMENT : Legal right to use the property of others for a specific purpose. For example, a utility  
company may have a five–foot easement along the property line of a home. This gives the utility the legal  
right to install and maintain a sewer line within the easement.  
EFFICIENCY : A ratio of total power output to the total power input, expressed as a percent.  
EFFLUENT : Wastewater or other liquid—raw (untreated), partially, or completely treated—flowing  
FROM a reservoir, basin, treatment process, or treatment plant.  
ELASTOMER : A natural or synthetic elastic material, often used for o–ring seals. Typical materials are  
viton, buna–n, EPDM (ethylene propylene dimonomer), etc.  
ELECTRO DEIONIZATION : Water treatment technology that utilizes an electricity, ion exchange mem-  
branes and resin to deionize water and separate dissolved ions (impurities) from water.  
ELEVATION : The height to which something is elevated, such as the height above sea level.  
ELUTRIATION : The washing of digested sludge with fresh water, plant effluent or other wastewater. The  
goal is to remove fine particles and/or the alkalinity in the sludge. This process reduces the demand for  
conditioning chemicals and improves settling or filtering characteristics of the sludge.  
ELUTRIATION : The washing of digested sludge with fresh water, plant effluent or other wastewater. The  
goal is to remove fine particles and/or the alkalinity in the sludge. This process reduces the demand for  
conditioning chemicals and improves settling or filtering characteristics of the sludge.  
EMERGENCY SEAT SEAL : To obtain tight shut off in an emergency situation, a sealant can be injected  
into a specially designed groove in the seat rings. Available for most ball valves and gate valves.  
EMULSION BREAKER (DEMULSIFIER) : Chemical additive that destroys the emulsifying characteris-  
tics water. Typically used separate stabilized (emulsified) oil in water.  
ENTHALPY : A thermodynamic property of a fluid. The enthalpy of a fluid consist of the energy associ-  
ated with the fluid at a microscopic level (related to the temperature of the fluid) plus the energy present in  
the form of pressure at the inlet and outlet of a system.  
EODD : EODD stands for Electrically Operated Double Diaphragm (pumps). These are diaphragm pumps  
driven either directly or indirectly with an electric motor. Offering many of the same advantages as  
AOD/AODD pumps with less noise and no compressed air/gas requirements. They cannot run against  
a closed discharge, which air operated models can, except Graco’s e–Series models.  
EPA : United States Environmental Protection Agency.  
497  
EQUALIZING BASIN : A holding basin in which variations in flow and composition of a liquid are  
averaged. Such basins are used to provide a flow of reasonably uniform volume and composition to a  
treatment unit. Also called a balancing reservoir.  
EQUALIZING BASIN : A holding basin in which variations in flow and composition of a liquid are  
averaged. Such basins are used to provide a flow of reasonably uniform volume and composition to a  
treatment unit. Also called a balancing reservoir.  
ESCHERICHIA COLIFORM : A species of bacteria found in large numbers in the intestinal tract of  
warm–blooded animals.  
ESDV : EMERGENCY SHUT DOWN VALVES – A valve or a system of valves which, when activated,  
initiate a shut–down of the plant, process, or platform they are tied to.  
ESTUARY : Bodies of water that are located at the lower end of a river and are subject to tidal fluctua-  
tions.  
ESTUARY : Bodies of water that are located at the lower end of a river and are subject to tidal fluctua-  
tions.  
EUTROPHICATION : The increase of nutrient levels of a lake or other body of water; this usually causes  
in increase in the growth of aquatic animal and plant life.  
EUTROPHICATION : The increase of nutrient levels of a lake or other body of water; this usually causes  
in increase in the growth of aquatic animal and plant life.  
EVAPOTRANSPIRATION : 1) The process by which water vapor passes into the atmosphere from living  
plants. Also called Transpiration. 2) The total water removed from an area by transpiration (living plants)  
and by evaporation from soil, snow and water surfaces.  
EVAPOTRANSPIRATION : 1) The process by which water vapor passes into the atmosphere from living  
plants. Also called Transpiration. 2) The total water removed from an area by transpiration (living plants)  
and by evaporation from soil, snow and water surfaces.  
EXFILTRATION : Liquid wastes and liquid–carried wastes which unintentionally leak out of a sewer pipe  
system and into the environment.  
EXPANDING GATE VALVE : A gate valve that is comprised of a separate gate and segment that as the  
valve operates the gate and segment move without touching the seats, permitting the valve to be opened  
and closed without wear. In the closed position the gate and segment are forced against the seat. Con-  
tinued downward movement of the gate causes the gate and segment to expand against the seats. When  
the valve reaches its full open position, the gate and segment seal off against the seats while the flow is  
isolated from the valve body.  
EXTENDED AERATION : A modification of the activated sludge process which provides for aerobic  
sludge digestion within the aeration system.  
EXTENDED BDV (BLOW DOWN VALVE) : Used on buried valves where the drain plug is inaccessi-  
ble. Instead, a line is piped above grade, terminating in a small valve. Line pressure is used to blowout  
condensates and other material which settles out in the bottom of the body cavity.  
EXTENSIONS : The equipment applied to buried valves to provide above grade accessibility to operating  
gear, blowdown and seat lubrication systems.  
FACE: TO–FACE – The overall dimension from the inlet face of a valve to the outlet face of the valve  
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Chapter 25. Glossary  
(one end to the other). This dimension is governed by ASME B16.10 and API–6D to ensure that such  
valves are mutually inter changeable, regardless of the manufacturer.  
FACULTATIVE : A combination of both aerobic and anaerobic conditions. Facultative cells have both aer-  
obic and anaerobic zones. Facultative bacteria are able to exist in both aerobic and anaerobic conditions.  
A facultative pond is commonly used to treat wastewater flows in small communities, It has an upper  
aerobic zone, a lower anaerobic zone, and algae provide most of the oxygen for the bacteria.  
FACULTATIVE ORGANISMS : Organisms that can survive and function in the presence or absence of  
free, elemental oxygen. Basically, organisms that can switch from aerobic or anaerobic depending on its  
environment. 227  
FACULTATIVE POND (ALSO KNOWN AS A WASTEWATER TREATMENT POND OR LAGOON)  
: The most common type of treatment pond used for treating domestic wastewater. The upper portion is  
aerobic, while the bottom layer is anaerobic. Algae supply most of the oxygen in the aerobic layer.  
FAIL SAFE VALVE : A valve designed to fail in a preferred position (open or closed) in order to avoid an  
undesirable consequence in a piping system.  
FAIR LEAD PULLEY : A pulley that is placed in a manhole to guide TV camera electric cables and the  
pull cable into the sewer when inspecting pipelines.  
FATS OIL & GREASE : a term to describe wastewater contaminants that are commonly found in food or  
petroleum based effluents. EPA test 1664 is typically used to measure FOG.  
FATS, OILS, & GREASE : Food industry byproducts that can cause significant problems for sewer sys-  
tems. This pollutant includes both animal/vegetable and petroleum sources and can be regulated sepa-  
rately by these fractions. It is important to know which fraction is regulated and what analytical method is  
being used to get accurate results.  
FECAL COLIFORM : A type of bacteria found in the bodily discharges of warm–blooded animals. Used  
as an indicator organism.  
FERMENTATION : A process of decomposition of organic solid materials by bacteria and other biologi-  
cal actions.  
FERRIC CHLORIDE : Metal salt (FeCl3) commonly used as an coagulant in water clarification and as  
etching agent in chemical–etching.  
FIELD SERVICEABLE : A statement indicating that normal repair of the valve or replacement of operat-  
ing parts can be accomplished in the field without return to the manufacturer.  
FILAMENTOUS : The property of growing in long strings, or filaments. Algae and bacteria have fila-  
mentous forms. Algae filaments can clog up equipment and be a nuisance in receiving waters. Bacterial  
filaments area common cause of bulking in activated sludge.  
FILAMENTOUS ORGANISMS : Organisms that grow in a thread or filamentous form. Common types  
are Thiothrix and Actinomycetes. A common cause of sludge bulking in the activated sludge process.  
FILTER PRESS : Solids dewatering device that uses pressure differential applied to sludge within a series  
of plates with filter clothes. The plates (with clothes in them) are arranged in a plate pack with the sludge  
filling chambers created by recesses within each plate. Filter presses are often called Plate and Frame or  
Recessed Chamber Filter Presses.  
FIRE GATE : A gate or ball valve which is positioned in a pipeline at the entrance to a compressor station.  
499  
This valve is closed in case of fire in the compressor station. Closing the valve prevents the gas in the  
pipeline from feeding the fire.  
FIRE SAFE : A statement associated with a valve design which is capable of passing certain specified  
leakage and operational tests after exposure to fire. Must be referenced to a particular specification.  
FLEXIBLE TUBE VALVE : A special valve using a flexible sleeve or tube which acts as the closure  
element. Pressure applied to the jacket space surrounding the outside of the tube, controls the opening and  
closing of the valve.  
FLOAT LINE : A length of rope or heavy twine attached to a float, plastic jug or parachute to be carried  
by the flow in a sewer from one manhole to the next. This is called “stringing the line” and is used for  
pulling through winch cables, such as for a bucket machine work or closed–circuit television work.  
FLOAT VALVE : A valve which automatically opens or closes as the level of a liquid changes. The valve  
is operated mechanically by a float which rests on the top of the liquid.  
FLOATING BALL : A ball valve having a non–trunnion mounted ball. The ball is free to float between  
the seat rings, and thus causes higher torques.  
FLOC : The agglomeration of smaller particles in a gelatinous mass that can be more easily removed  
from the liquid than the individual small particles. Clumps of bacteria and particles or coagulants and  
impurities that have come together and formed a cluster. Found in aeration tanks, secondary clarifiers and  
chemical precipitation processes.  
FLOCCULANT : high molecular weight polyelectrolyte water soluble organic polymer designed to  
agglomerate solids in water substrates. Characteristics of flocculants in water treatment are determined by  
their molecular weight, charge type (anionic, nonionic or cationic) and charge density.  
FLOCCULATION : the agglomeration of settleable solids through a bridging mechanism to produce  
larger particle that is more easily separated from water.  
FLOODED SUCTION : In a flooded suction system, the liquid flows to the pump inlet from an elevated  
source by means of gravity. This is generally recommended for centrifugal pumps.  
FLOTATION : (1) The stress or forces on a pipeline or manhole structure below a water table which tend  
to lift or float the pipeline or manhole structure. (2) The process of raising suspended matter to the surface  
of the liquid in a tank where it forms a scum layer that can be removed by skimming. The suspended  
matter is raised by aeration, the evolution of gas, the use of chemicals, electrolysis, heat or bacterial  
decomposition.  
FLOW : A measure of the liquid volume capacity of a pump. Given in gallons per minute/hour (gpm,  
gph), liters per minute/hour (L/min, L/hour), milliliters per minute (mL/min), cubic meters per hour  
(m3/h) and other rarely used measurements.  
FLOW : The continuous movement of a liquid from one place to another.  
FLOW ISOLATION : A procedure used to measure inflow and infiltration (I/I). A section of sewer is  
blocked off or isolated and the flow from the section is measured.  
FLUME : (1) An open conduit of wood, masonry, metal, or plastic constructed on a grade and sometimes  
elevated. (2) A flow rate measurement device.  
FLUSHER BRANCH : A line built specifically to allow the introduction of large quantities of water to  
the collection system so the lines can be “flushed out” with water. Also installed to provide access for  
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Chapter 25. Glossary  
equipment to clear stoppages in a sewer.  
FLUSHING : The removal of deposits of material which have lodged in sewers because of inadequate ve-  
locity of flows. Water is discharged into the sewers at such rates that the larger flow and higher velocities  
are sufficient to remove the material.  
FLUX : the permeate rate per unit area of membrane surface. Typical units are gals per foot of membrane  
per day (gfd) or liters of permeate per square meter of membrane per hour (lmh).  
FOOD: TO–MASS RATIO (F/M) – An activated sludge process–control calculation based upon the  
amount of food (BOD5 or COD) available per pound of mixed liquor volatile suspended solids.  
FORCE MAIN : A pipe that carries wastewater under pressure from the discharge side of a pump to a  
point of gravity flow downstream.  
FREE AVAILABLE RESIDUAL CHLORINE : That portion of the total available chlorine residual com-  
posed of dissolved chlorine gas (Cl2), hypochlorous acid (HOCl), and/or hypochlorite ion (OCl–) remain-  
ing in water after chlorination.  
FREE OXYGEN : Oxygen can be dissolved in water as the soluble gas O2 when it is called free oxygen  
and measured as dissolved oxygen.  
FREEBOARD : The vertical distance from the normal water surface to the top of the confining wall.  
FRICTION : The force produced as a reaction to movement. All fluids produce friction when they are  
in motion. The higher the fluid viscosity, the higher the friction force for the same flow rate. Friction  
is produced internally as one layer of fluid moves with respect to another and also at the fluid/surface  
interface.  
FRICTION HEAD : The pressure expressed in pounds per square inch or feet of liquid needed to over-  
come the resistance to the flow in pipes and fittings.  
FRICTION HEAD DIFFERENCE : The difference in head required to move a mass of fluid from one  
position to another at a certain flow rate.  
FRICTION LOSS : The head lost by water flowing in a stream or conduit as the result of the disturbances  
set up by the contact between the moving water and its containing conduit and by intermolecular friction.  
FULL BORE (FULL PORT) : Describes a valve in which the bore (port) is nominally equal to the bore of  
the connecting pipe.  
FULL OPENING : Describes a valve whose bore (port) is nominally equal to the bore of the connecting  
pipe.  
GAC : Granular Activated Carbon – A material used to absorb organics from wastewater. This char-  
coal–like material can be used in filtration systems to remove solvent contamination.  
GATE : The closure element of a gate valve (sometimes called wedge or disc).  
GATE VALVE : A straight–thru pattern valve whose closure element is a wedge or parallel–sided slab, sit-  
uated between two fixed seating surfaces, with means to move it in or out of the flow stream in a direction  
perpendicular to the pipeline axis.  
GLAND FOLLOWER OR GLAND FLANGE : The component used to hold down or retain the gland in  
the stuffing box.  
GLAND OR GLAND BUSHING : That part of a valve which retains or compresses the stem packing in a  
stuffing box (where used) or retains a stem O–ring, lip seal, or stem O–ring bushing. Sometimes manually  
501  
adjustable.  
GLAND PLATE : The plate in a valve which retains the gland, gland bushing or stem seals and some-  
times guides the stem.  
GLOBE VALVE : A valve whose closure element is a flat disc or conical plug sealing on a seat which is  
usually parallel to the flow axis. Can be used for throttling services.  
GO : GEAR OPERATED – The actuation of a valve thru a – ear set which multiplies the torque applied to  
the valve stem.  
GRAB SAMPLE : A single sample of water collected at a particular time and place which represents the  
composition of water only at that time and place.  
GRADE : (1) The elevation of the invert (or bottom) of a pipeline, canal, culvert, sewer, or similar conduit.  
(2) The inclination of slope of a pipeline, conduit, stream channel, or natural ground surface; usually ex-  
pressed in terms of the ratio or percentage of number of units of vertical rise or fall per unit of horizontal  
distance. A 0.5 percent grade would be a drop of one–half foot per hundred feet of pipe.  
GRAVITY FLOW : Water or wastewater flowing from a higher elevation to a lower elevation due to  
the force of gravity. The water does not flow due to energy provided by a pump. Wherever possible,  
wastewater collection systems are designed to use the force of gravity to convey waste liquids and solids.  
GREASE : In wastewater, a group of substances, including fats, waxes, free fatty acids, calcium and  
magnesium soaps, mineral oils, and certain other non–fatty materials.  
GREASE : In a collection system, grease is considered to be the residues of fats, detergents, waxes, free  
fatty acids,  
GREASE BUILDUP : Any point in a collection system where coagulated and solidified greases accu-  
mulate and build up. Many varieties of grease have high adhesive characteristics and collect other solids,  
forming restrictions and stoppages in collection systems.  
GREASE FITTING : A fitting through which lubricant or sealant is injected.  
GREASE TRAP : A receptacle designed to collect and retain grease and fatty substances usually found in  
kitchens or from similar wastes. It is installed in the drainage system between the kitchen or other point  
of production of the waste and the building wastewater collection line. Commonly used to control grease  
from restaurants.  
GREEN ALGAE : The common forms of algae in an aerobic lagoon environment. Green algae are essen-  
tial for lagoon treatment.  
GRINDER PUMP : A grinder pump is a waste management device. Waste from water–using household  
appliances (toilets, bathtubs, washing machines, etc.) flows through the home’s pipes into the grinder  
pump’s holding tank. Once the waste inside the tank reaches a certain level, the pump will turn on, grind  
the waste into fine slurry, and pump it to the central sewer system.  
GRIT : The heavy mineral material present in wastewater such as sand, coffee grounds, eggshells, gravel  
and cinders. Grit tends to settle out at flow velocities below 2 ft. /sec,  
GRIT CATCHER : A chamber usually placed at the upper end of a depressed collection line or at other  
points on combined or storm water collection lines where wear from grit is possible. The chamber is sized  
and shaped to reduce the velocity of flow through it and thus permit the settling out of grit.  
GRIT REMOVAL : Grit removal is accomplished by providing an enlarged channel or chamber which  
502  
Chapter 25. Glossary  
causes the flow velocity to be reduced and allows the heavier grit to settle to the bottom of the channel  
where it can be removed.  
GRIT REMOVAL : Grit removal is accomplished by providing an enlarged channel or chamber which  
causes the flow velocity to be reduced and allows the heavier grit to settle to the bottom of the channel  
where it can be removed.  
GRIT TRAP : A permanent structure built into a manhole (or other convenient location in a collection  
system) for the accumulation and easy removal of grit.  
HAND WHEEL : A wheel–shaped valve operating device intended to be grasped with one or both hands  
which allows turning the valve stem or operator shaft to which it is attached.  
HARD FACING : A surface preparation in which an alloy is deposited on a metal surface usually by weld  
overlay to increase resistance to abrasion and or corrosion.  
HARD WATER : Water having a high concentration of calcium and magnesium ions.  
HARDNESS : It is typically the concentration of calcium and magnesium salts in water. However, it may  
include other metal salts such as Al, Mn, Sr and Zn. Normally measured as CaCO3 equivalents.  
HEAD : Refers to the pressure produced by a vertical column of fluid. It is a measure of pressure, ex-  
pressed in feet of head for pumps. Water is used as the default where 10 meters (33.9 ft.) of water equals  
one atmosphere (14.7 psi. or 1 bar).  
HEAD : The vertical distance (in feet) equal to the pressure (in psi) at a specific point. The pressure head  
is equal to the pressure in psi times 2.31 ft/psi.  
HEADWORKS : The part of wastewater treatment where the raw sewage is first received for treatment.  
Preliminary treatment occurs at the Headworks.  
HETEROTROPH : a type of bacteria that uses organic matter for energy and can use free oxygen, nitrates  
or sulfates as and oxygen source for respiration.  
HUBS : The end connection tubes on a gate valve.  
HUMAN MACHINE INTERFACE (HMI) : An interface such as a display screen that permits interaction  
between a human being and a device/equipment or process. HMI allows for the visual display of status  
and condition of a particular process or equipment.  
HWO : HANDWHEEL OPERATED – A valve on which the handwheel drives the stem directly to oper-  
ate the valve.  
HYDRAULIC LOADING : The flow of water per acre of surface area.  
HYDRAULIC MOTOR ACTUATOR (OPERATOR) : A device by which rotation of a hydraulically  
powered motor is converted into mechanical motion.  
HYDROGEN SULFIDE : A very odorous and poisonous gas. Commonly known as rotten egg gas. It is a  
combined form of hydrogen and sulfur with the formula H2S.  
HYDROGEN SULFIDE GAS (H2S) : Hydrogen sulfide is a gas with a rotten egg odor. This gas is pro-  
duced under anaerobic conditions. Hydrogen sulfide is particularly dangerous because it dulls your sense  
of smell so that you do not notice it after you have been around it for a while and because the odor is  
not noticeable in high concentrations. The gas is very poisonous to your respiratory system, explosive,  
flammable and colorless.  
HYDROLOGIC CYCLE : The process of evaporation of water into the air and its return to earth by  
503  
precipitation. (Also called the WATER CYCLE)  
HYPOCHLORINATORS : Chlorine pumps, chemical feed pumps or devices used to dispense chlorine  
solutions made from hypochlorites into the water being treated.  
IMPELLER : The rotating element of a centrifugal pump which imparts movement and pressure to a fluid.  
INCLINED PLATE CLARIFIER : a solids / liquid separation device (clarifier) that is filled with parallel  
(sometimes called Lamella) plates that are inclined at an angle between 45 and 55 degrees. The plates  
reduce (compared to a gravity clarifier) the foot–print required to properly settle solids.  
INCREMENTAL SEAT TEST : The leakage testing of valve seats in an assembled valve by increasing the  
applied pressure in prescribed pressure steps.  
INDUSTRIAL WASTEWATER : Wastes associated with industrial manufacturing processes. 229  
INFECTION : Introduction of presence of pathogenic organisms in potable water supply.  
INFECTION : Introduction of the presence of pathogenic organisms in potable water supply. This is  
determined in two ways:  
INFILTRATION : The gradual flow of water into the soil; also, the flow of groundwater as seepage into a  
sewer system.  
INFILTRATION HEAD : The distance from a point of infiltration leaking into a collection system to  
the water table elevation. This is the pressure of the water being forced through the leak in the collection  
system.  
INFILTRATION/INFLOW : The total quantity of water from both infiltration and inflow without distin-  
guishing the source. Abbreviated I&I or I/I.  
INFLATABLE PIPE STOPPER : An inflatable ball or bag used to form a plug to stop flows in a sewer  
pipe.  
INFLOW : Water discharged into a sewer system and service connections from sources other than regular  
connections. This includes flow from yard drains, foundation drains and around manhole covers. Inflow  
differs from infiltration in that it is a direct discharge into the sewer rather than a leak in the sewer itself.  
INFLUENT : Wastewater or other liquid—raw (untreated) or partially treated—flowing into a reservoir,  
basin, treatment process, or treatment plant.  
INLET : (1) A surface connection to a drain pipe. (2) A chamber for collecting storm water with no well  
below the outlet pipe for collecting grit. Often connected to a CATCH BASIN or a “basin manhole” with  
a grit chamber.  
INLET PORT : That end of a valve which is connected to the upstream pressure zone of a fluid system.  
INNER SEAT RING : The inner part of a two–piece valve seat assembly.  
INORGANIC : Material such as salts, metals, and all other substances of mineral origin.  
INORGANIC : Material such as sand, salt, iron, calcium salts and other mineral materials and other than  
of plant or animal origin or of carbon compounds (ORGANIC).  
INORGANIC MATERIAL : Material that will not respond to biological action (sand, cinders, stone).  
Non–volatile fraction of solids.  
INORGANIC MATERIAL : Material that will not respond to biological action (sand, cinders, stone).  
Non–volatile fraction of solids.  
504  
Chapter 25. Glossary  
INSERTION PULLER : A device used to pull long segments of flexible pipe material into a sewer line  
when sliplining to rehabilitate a deteriorated sewer.  
INSIDE: OUT AIR SEAT TEST – A pressure test that can be performed only on independent seating  
trunnion mounted ball valves. By closing the valve and pressurizing the body cavity, all of the seals in an  
independent seating ball valve can then be pressure tested.  
INSITUFORM : A method of installing a new pipe within an old pipe without excavation. The process  
involves the use of a polyester–fiber felt tube, lined on one side with polyurethane and fully impregnated  
with a liquid thermal setting resin.  
INSPECTION TELEVISION EQUIPMENT : Television equipment that is superior to standard commer-  
cial quality, providing 600 to 650 lines of resolution, and designed for industrial inspection applications.  
INTEGRATED FIXED FILM ACTIVATED SLUDGE (IFAS) : A suspended growth system that pro-  
vides additional biomass within a biological wastewater treatment system to meet more stringent effluent  
parameters or increased loadings.  
INTERNAL PRESSURE RELIEF : A self relieving feature in non–independent seating valves that auto-  
matically relieves excessive internal body pressure caused by sudden changes in line pressures. By means  
of the piston effect principal the excessive body pressure will move the seat away from its seating surface  
and relieve it to the lower pressure side.  
INVERT : The lowest point of the channel inside a pipe or manhole.  
INVERTED SIPHON : A pressure pipeline used to carry wastewater flowing in a gravity collection sys-  
tem under a depression such as a valley or roadway or under a structure such as a building. 230  
ION EXCHANGE : Process that removes dissolved ions from solution of a certain charge by absorption  
onto a resin that releases (exchanges) an ion of the same charge.  
ISRS : Inside screw, rising stem – common term for any valve design in which the stem threads are ex-  
posed to the fluid below the packing and the stem rises up through the packing when the valve is opened.  
KEY MANHOLE : In collection system evaluation, a key manhole is one from which reliable or specific  
data can be obtained.  
KEY STOP : A method of restricting the travel of a ball valve from fully open to fully closed. The stem  
key bears against the ends of an arc machined in the adaptor plate.  
KITE : A device for hydraulically cleaning sewer lines. Resembling an airport wind sock and constructed  
of canvas–type material, the kite increases the velocity of a flow at its outlet to wash debris ahead of it.  
LAMINAR : A distinct flow regime that occurs at low Reynolds number (Re < 2000). It is characterized  
by particles in successive layers moving past one another in a well behaved manner (little to no turbu-  
lence).  
LAMPING : Using reflected sunlight or a powerful light beam to inspect a sewer between two adjacent  
manholes. The light is directed down the pipe from one manhole. If it can be seen from the next manhole,  
it indicates that the line is open and straight.  
LATERAL : (See LATERAL SEWER)  
LATERAL CLEANOUT : A capped opening in a building lateral, usually located on the property line,  
through which the pipelines can be cleaned.  
LATERAL SEWER : A sewer that discharges into a branch or other sewer and has no other common  
505  
sewer tributary to it. Sometimes called a “street sewer” because it collects wastewater from individual  
homes.  
LEAD : A heavy metal commonly regulated by wastewater discharge permits. Other heavy metals include  
– Arsenic (As), Cadmium (Cd), Chromium(Cr), Copper (Cu), Nickel (Ni), and Zinc (Zn).  
LEVER : A handle type operating device for quarter–turn valves.  
LIFT STATION : A wastewater pumping station that lifts the wastewater to a higher elevation when  
continuing the sewer at reasonable slopes would involve excessive depths of trench. Also, an installation  
of pumps that raise wastewater from areas too low to drain into available sewers. These stations may be  
equipped with air–operated ejectors or centrifugal pumps. Sometimes called a PUMP STATION, but this  
term is usually reserved for a similar type of facility that is discharging into a long FORCE MAIN, while a  
lift station has a discharge line or force main only up to the downstream gravity sewer.  
LIFTING LUGS : Lugs provided on large ball, gate, and check valves, for lifting and positioning valves.  
Also called lifting eyes.  
LIMIT SWITCH : An electrical device providing a signal to a remote observation station indicating when  
the valve is in the fully open or fully closed position. Usually a component of a valve operator.  
LIQUID END : A term referring to the side (or parts) of the pump that come into contact with the process  
fluid. This applies to any air operated pump including Air Operated Diaphragm Pumps, air operated piston  
pumps and air operated drum pumps.  
MAGNETIC DRIVE : Also referred to as a Mag–drive. This is a method of connecting the motive force  
to the pump which uses a series of magnets coupled together, with a containment chamber separating  
them. Magnetic Drives keep the fluid sealed from atmosphere and other environmental factors and elim-  
inate the need for seals and seal maintenance. Special considerations must be taken into account when  
specifying a Mag–Drive pump or mixer. Ask Springer Pumps for more information.  
MAIN LINE : Branch or lateral sewers that collect wastewater from building sewers and service lines.  
MAIN SEWER : A sewer line that receives wastewater from many tributary branches and sewer lines and  
serves as an outlet for a large territory or is used to feed an intercepting sewer.  
MANDREL : (1) A special tool used to push bearings in or to pull sleeves out. (2) A gage used to measure  
for excessive deflection in a flexible conduit.  
MANHOLE : An opening in a sewer provided for the purpose of permitting operators or equipment to  
enter or leave a sewer.  
MANHOLE ELEVATION : The height (elevation) of the invert or lowest point in the bottom of a manhole  
above mean sea level.  
MANHOLE FLOW : (1) The depth or amount of wastewater flow in a manhole as observed at any se-  
lected time. (2) The total or the average flow through a manhole in gallons on any selected time interval.  
231  
MANHOLE INFILTRATION : Groundwater that seeps or leaks into a manhole structure.  
MANHOLE INFLOW : Surface waters flowing into a manhole, usually through the vent holes in the  
manhole lid.  
MANHOLE INVERT : The lowest point in a trough or flow channel in the bottom of a manhole.  
MANHOLE LID : The heavy cast–iron or forged–steel cover of a manhole. The lid may or may not have  
506  
Chapter 25. Glossary  
vent holes.  
MANHOLE LID DUST PAN : A sheet metal or cast–iron pan located under a manhole lid. This pan  
serves to catch and hold pebbles and other debris falling through vent holes, preventing them from getting  
into the pipe system.  
MANHOLE VENTS : One or a series of one–inch diameter holes through a manhole lid for purposes of  
venting dangerous gases found in sewers.  
MASKING AGENTS : Substances used to cover up or disguise unpleasant odors. Liquid masking agents  
are dripped into the wastewater, sprayed into the air, or evaporated (using heat) with the unpleasant fumes  
or odors and then discharged into the air by blowers to make an undesirable odor less noticeable.  
MASKING AGENTS : Substances used to cover up or disguise unpleasant odors. Liquid masking agents  
are dripped into the wastewater, sprayed into the air, or evaporated (using heat) with the unpleasant fumes  
or odors and then discharged into the air by blowers to make an undesirable odor less noticeable.  
MBR : Membrane Bio Reactor – A wastewater treatment technology that combines biological treatment  
with physical treatment involving membrane filtration. It’s used primarily to treat BOD, COD, and sus-  
pended solids to very low levels where effluent may be able to be reused or recycled.  
MEAN CELL RESIDENCE TIME (MCRT) : The average length of time a mixed liquor suspended solids  
particle remains in the activated sludge process. May also be known as sludge retention time.  
MECHANICAL AERATION : Method of aeration.  
MECHANICAL AERATION : The use of machinery to mix air and water so that oxygen can be absorbed  
into the water.  
MECHANICAL CLEANING : Clearing pipe by using equipment that scrapes, cuts, pulls or pushes  
the material out of the pipe. Mechanical cleaning devices or machines include bucket machines, power  
rodders and hand rods.  
MECHANICAL PLUG : A pipe plug used in sewer systems that is mechanically expanded to create a  
seal.  
MECHANICAL SEAL : In a valve, a shut off that is accomplished by a mechanical means rather than  
with fluid or line pressure. The wedging action of a gate against the seats or the seat springs pushing the  
seat against the ball or gate are examples of mechanical seals in a valve.  
MEMBRANE : Material layer that is a selective barrier between two phases and remains impermeable to  
specific particles, molecules or substances.  
MEMBRANE BIO REACTOR : Aerobic biological wastewater treatment process that utilizes membrane  
filtration (rather than clarification) for solids / liquid separation. The membrane filters (ultra filters) can be  
either submerged or external to the biological mixed liquor tanks.  
MEMBRANE BIOREACTOR (MBR) : Biological wastewater treatment process where a selected mem-  
brane is integrated with a biological process to act as a suspended growth bioreactor.  
MERCURY (HG) : A metal which remains liquid at room temperature. This property makes it useful  
when used in a thin vertical glass tube since small changes in pressure can be measured as changes in the  
mercury column height. The inch of mercury is often used as a unit for negative pressure.  
METAL: TO–METAL SEAL – The seal produced by metal–to–metal contact between the sealing face of  
the seat ring and the closure element, without benefit of a synthetic seal.  
507  
METERING PUMP : Pumps used for precise introductions of chemicals into a tank, existing fluid stream  
or some other liquid handling equipment. Types of pumps for these include Diaphragm Pumps (AOD  
or EOD), Peristaltic Pumps, Hose Pumps, Gear Pumps, Bellows Pumps, Piston Pumps and other less  
commonly used pump types.  
METHANE : A combustible gas produced during anaerobic fermentation of organic matter, such as by  
anaerobic digestion of wastewater solids.  
MGO : MANUAL GEAR OPERATOR – A gear operator that is operated manually (with a handwheel).  
MGWPCS PERMIT : Montana Groundwater Pollution Control System permit. This permit is issued to  
owners/operators of potential sources of pollution to state ground waters.  
MICRO FILTRATION : Type of membrane filtration that separates suspended solids and solutes of high  
molecular from a liquid and low molecular weight solutes. This separation process is used in industry  
processes, water treatment and research for purifying and concentrating macromolecular solutions.  
MICRO: Organisms – Microscopic plans and animals such as bacteria, molds, protozoa, algae, and small  
metazoa.  
MICRO: Organisms – Microscopic plants and animals such as bacteria, molds, protozoa, algae, and small  
metazoa.  
MICROORGANISMS : Microscopic living organisms.  
MICROORGANISMS : Very small organisms that can be seen only through a microscope. Some microor-  
ganisms use the wastes in wastewater for food and thus remove or alter much of the undesired matter.  
MILLIGRAMS PER LITER, MG/L : A measure of the concentration by weight of a substance per unit  
volume. One thousandth of a gram in one liter. One mg/L is equal to one part per million (ppm).  
MIXED BED ION EXCHANGE : Anion exchange process that uses a mixture of cation and anion resin  
combined in a single ion exchange column. With proper pretreatment, product water purified from a  
single pass through a mixed bed ion exchange column is the purest that can be made.  
MIXED LIQUOR : When the activated sludge in an aeration tank is mixed with primary effluent or the  
raw wastewater and return sludge, this mixture is then referred to as mixed liquor as long as it is in the  
aeration tank. Mixed liquor may also refer to the contents of mixed aerobic or anaerobic digesters.  
MIXED LIQUOR SUSPENDED SOLIDS (MLSS) : The milligrams of suspended solids per liter of  
mixed liquor that are combustible at 550 degrees Centigrade. An estimate of the quantity of MLSS to be  
wasted from the aeration tank of an extended aeration plant may be determined by the rate of settling and  
centrifuge tests on the sludge solids.  
MIXED LIQUOR VOLATILE SUSPENDED SOLIDS (MLSS) : The concentration of organic matter in  
the mixed liquor suspended solids.  
MIXED LIQUOR VOLATILE SUSPENDED SOLIDS (MLVSS) : The organic or volatile suspended  
solids in the mixed liquor of an aeration tank. This volatile portion is used as a measure or indication of  
the microorganisms present.  
MIXED MEDIA GRAVITY FILTER : A filter using more than one filtering media (such as coal and  
sand).  
MIXER, SUBMERSIBLE : A submersible mixer is a mechanical device that is used to mix sludge tanks  
and other liquid volumes. Submersible mixers are often used in sewage treatment plants to keep solids in  
508  
Chapter 25. Glossary  
suspension in the various process tanks and/or sludge holding tanks. The submersible mixer is operated  
by an electric motor, which is coupled to the mixer’s propeller, either direct–coupled or via a planetary  
gear–reducer. The propeller rotates and creates liquid flow in the tank, which in turn keeps the solids in  
suspension. The submersible mixer is typically installed on a guide rail system, which enables the mixer  
to be retrieved for periodic inspection and preventive maintenance.  
MIXER, VERTICAL : This style of mixer uses an extended shaft between the motor and mixing blade  
such that the motor is above and out of the liquid. These also incorporate a gear reducer to slow the  
speed of the mixing blades to achieve the desired mixing/tank turnover rate. Vertical mixers are often  
used in reactor vessels to ensure thorough chemical reactions or to mix different ingredients together in  
food/beverage applications.  
MOISTURE CONTNET : The amount of water per unit weight of bio solids.  
MONITORING WELL : Wells used to collect groundwater samples for analysis to determine the amount,  
type, and spread of contaminants in groundwater. Specific design is often determined by the Water Protec-  
tion Bureau at MT DEQ.  
MOVING BED BIO REACTOR : Aerobic biological wastewater treatment process that utilizes the fixed  
film (media) process. High surface area media is suspended in biological mixed liquor and bacteria grow  
on the media (attached growth) surface. A clarifier is typically used downstream for solid / liquid separa-  
tion.  
MOVING BED BIOREACTOR (MBBR) : Method to biologically treat wastewater by circulating moving  
media in an aerobic sludge environment.  
MPDES PERMIT : Montana Pollutant Discharge Elimination System permit. This permit lists the condi-  
tions that must be met before treatment plants can discharge an effluent into state  
MULTI: Stage Pump – The multi–stage pump is used for clean, clear liquids requiring significant dis-  
charge pressure. A multi–stage pump is nothing more than a standard centrifugal pump with the discharge  
of the initial volute discharging directly into the suction of the next volute. The numerous “volutes” are  
all internal to the pump and many times the volutes are hard to spot individually. The number of stages is  
dependent on the desired Total Discharge Head required by the application. These types of pump can be  
either horizontal or vertical configuration. Commonly used for boiler feed.  
NEEDLE VALVE : A type of small valve, used for flow metering, having a tapered needle–point plug or  
closure element and a seat having a small orifice.  
NEGATIVE PRESSURE : Pressure that is less than the pressure in the external environment.  
NEPHELOMETER TURBIDITY UNITS : It is a measurement of the clarity (turbidity) of a liquid.  
NEPHELOMETRIC TURBIDITY UNIT : The standard unit to measure turbidity or how cloudy water is.  
It’s used as a visual indicator for how well a wastewater treatment system is working.  
NET POSITIVE SUCTION HEAD (NPSH.) : The head in feet of water absolute as measured or cal-  
culated at the pump suction flange, less the vapor pressure (converted to feet of water absolute) of the  
fluid.  
NET POSITIVE SUCTION HEAD AVAILABLE (NPSHA) : The NPSHa available to prevent cavita-  
tion of the pump. To calculate the NPSHa, take the (Static Suction Head) plus (Suction Vessel Surface  
Pressure Head) minus (vapor pressure of fluid) minus (friction losses in the suction).  
NET POSITIVE SUCTION HEAD REQUIRED (NPSHR) : The NPSHr to stop a pump from cavitating.  
509  
The NPSHr is generally supplied to you by the pump manufacturer.  
NEWTONIAN FLUID : A fluid where the relation between shear stress and shear rate is linear, related to  
viscosity.  
NICKEL : A heavy metal commonly regulated by wastewater discharge permits. It can be found in met-  
als–related industries and products, including stainless steel. Other heavy metals include – Arsenic (As),  
Cadmium (Cd), Chromium(Cr), Copper (Cu), Lead (Pb), and Zinc (Zn).  
NITRIFICATION : The conversion of nitrogen matter into nitrates by bacteria.  
NITRIFICATION : An aerobic process in which bacteria change ammonia and organic nitrogen into  
nitrite and nitrate forms of nitrogen.  
NITRIFICATION : Wastewater treatment process involved in the biological removal of ammonia in which  
the ammonia is converted to nitrates (NO3)  
NITRIFYING BACTERIA : Bacteria that change the ammonia and organic nitrogen in wastewater into  
oxidized nitrogen (usually nitrate).  
NITROGEN : Nitrogen is present in wastewater in many forms – total Kjeldahl nitrogen, ammonia nitro-  
gen, organic nitrogen.  
NITROGEN CYCLE : The cycle of life, death, and decay involving organic nitrogenous matter is known  
as the nitrogen cycle. In the nitrogen cycle ammonia is produced from proteins.  
NON: Newtonian Fluid – A fluid with properties that is different in any way from those of Newtonian  
Fluids. This is usually found in the relation of viscosity and shear or shear/time.  
NON: RISING STEM – A gate valve having its stem threaded into the gate. As the stem turns, the gate  
moves, but the stem does not rise. Stem threads are exposed to line fluids.  
NON: Wetted Parts – A term used any part of a pump, or other type of liquid handling equipment, that  
doesn’t come into direct contact with the process fluid.  
NONIONIC : Neutral charged high molecular weight polyelectrolyte water soluble organic polymer  
designed to agglomerate solids in water substrates.  
NONPOINT DISCHARGE : A source of wastewater that comes from a relatively large area and would  
have to be controlled by a management or conservation practice. Storm waters and most agricultural  
waters are nonpoint sources.  
NONPOTABLE : Water that is considered unsafe and/or unpalatable for drinking.  
NONTRANSIENT NONCOMMUNITY (NTNC) WATER SYSTEM : Means a public water system that  
is not a community water system and that regularly serves at least 25 of the same persons over six months  
per year, including schools, day care centers, factories, restaurants and hospitals. 232  
NORMALLY CLOSED SOLENOID VALVE : An electrically operated valve whose inlet orifice is closed  
when the solenoid coil is not energized. Energize to open.  
NPDES PERMIT : National Pollutant Discharge Elimination System permit is the regulatory agency doc-  
ument issued by either a federal or state agency which is designed to control all discharges of pollutants  
from all point sources and storm water runoff into U.S. waterways. A treatment plant that discharges to a  
surface water will have a NPDES permit.  
NRS : NON–RISING STEM – A gate valve having its stem threaded into the gate. As the stem turns the  
gate moves but the stem does not rise. Stem threads are exposed to the line fluid.  
510  
Chapter 25. Glossary  
NUTRIENTS : Substances required by living plants and organisms. Forms of nitrogen and phosphorous  
are nutrients that can cause problems in receiving waters.  
OBSTRUCTION : Any solid object in or protruding into a wastewater flow in a collection line that pre-  
vents a smooth or even passage of the wastewater.  
OFFSET : (1) A combination of elbows or bends which brings one section of a line of pipe out of line  
with, but into a line parallel with, another section. (2) A pipe fitting in the approximate form of a reverse  
curve, made to accomplish the same purpose. (3) A pipe joint that has lost its bedding support and one  
of the pipe sections has dropped or slipped, thus creating a condition where the pipes no longer line up  
properly.  
OPDES : Oklahoma Pollutant Discharge Elimination System – A permit program established in accor-  
dance with Section 402 of the CWA and authorized in 27A O.S. Environment and Natural Resources. This  
program regulates discharges into Oklahoma’s waters from point sources, including municipal, industrial,  
commercial and certain agricultural sources.  
OPERATING POINT : The point on the system curve corresponding to the flow and head required to  
meet the process requirements.  
OPERATOR : A device which converts manual, hydraulic, pneumatic or electrical energy into mechanical  
motion to open and close a valve.  
ORGANIC : Material which comes from mainly animal or plant sources and contains carbon.  
ORGANIC ACIDS : Weak acids formed from organic compounds, such as acetic acid and citric acid.  
These acids form first in anaerobic digesters and then are converted to methane. The organic acids in  
wastewater lagoons are much more complex and generally weaker.  
ORGANIC MATERIAL : Carbon based material that can be broken down by bacteria (fats, meats, plant  
life). Represents the volatile fraction of solids.  
ORTHOPHOSPHATE : A simple compound of phosphorous and oxygen that is soluble in water.  
OSANDY : OUTSIDE SCREW AND YOKE – A valve in which the fluid does not come in contact with  
the stem threads. The stem sealing elements is between the valve body and the stem threads.  
OUTFALL : (1) The point, location or structure where wastewater or drainage discharges from a sewer,  
drain, or other conduit. (2) The conduit leading to the final disposal point or area.  
OUTFALL SEWER : A sewer that receives wastewater from a collection system or from a wastewater  
treatment plant and carries it to a point of ultimate or final discharge in the environment.  
OUTLET : Downstream opening or discharge end of a pipe, culvert, or canal.  
OVERFLOW MANHOLE : A manhole which fills and allows raw wastewater to flow out onto the street  
or ground.  
OVERFLOW RELIEF LINE : Where a system has overload conditions during peak flows, an outlet may  
be installed above the invert and leading to a less loaded manhole or part of the system. This is usually  
called an “overflow relief line.”  
OXIC : A biological environment which is aerobic.  
OXIDATION : Oxidation is the addition of oxygen, removal of hydrogen, or the removal of electrons from  
an element or compound. In wastewater treatment, organic matter is oxidized to more stable substances.  
OXIDATION : Oxidation is the addition of oxygen, removal of hydrogen, or the removal of electrons from  
511  
an element or compound. In wastewater treatment, organic matter is oxidized to more stable substances.  
OXIDATION POND : A term often used interchangeably with lagoon. Oxidation ponds are used after  
other treatment processes.  
OXIDATION PONDS OR LAGOONS : Holding ponds designed to allow the decomposition of organic  
wastes by aerobic or anaerobic means.  
OXIDATION PONDS OR LAGOONS : Holding ponds designed to allow the decomposition of organic  
wastes by aerobic or anaerobic means.  
OXYGEN REDUCTION POTENTIAL : A measure that indicates the capacity of wastewater to gain  
or reduce electrons during a chemical reaction. It is used as a control parameter for treating hexavalent  
chromium wastewater in the metal finishing industry.  
PACKAGE TREATMENT PLANT : A small wastewater treatment plant often fabricated at the manufac-  
turer’s factory, hauled to the site, and installed as one facility. The package may be either a small primary  
or a secondary wastewater treatment plant.  
PACKING : The deformable sealing material inserted into a valve stem stuffing box, which, when com-  
pressed by a gland, provides a tight seal about the stem.  
PALMER: BOWLUS FLUME – A flow measuring device consisting of a preformed flume.  
PARACHUTE : A device used to catch wastewater flow to pull a float line between manholes.  
PARSHALL FLUME : A flow measuring device consisting of a preformed flume with restrictive area  
called the throat. The head of water at a stilling well just upstream from the narrow part of the throat is  
measured and a chart is used to obtain flow rate.  
PARTS PER BILLION : A unit of concentration for pollutants in the wastewater. It’s the equivalent of one  
microgram in 1 liter (ug/l).  
PARTS PER MILLION : A unit of concentration for pollutants in the wastewater. It’s the equivalent of 1  
milligram in 1 liter (mg/l).  
PATHOGENIC ORGANISMS : bacteria, viruses or cysts, which can cause disease (typhoid, cholera,  
dysentery) in a host such as a human. Also called Pathogens.  
PEAKING FACTOR : Ratio of a maximum flow to the average flow, such as maximum hourly flow or  
maximum daily flow to the average daily flow.  
PERCOLATION : The movement or flow of waster through soil or rocks.  
PERCOLATION : The movement or flow of waster through soil or rocks.  
PERCOLATION : The movement or flow of water through soil or rocks. A discharge option for many  
wastewater treatment systems. In Montana, a Montana Ground Water Pollution Control System (MGW-  
PCS) permit and sampling in groundwater monitoring wells are required for permitted systems.  
PERFORMANCE CURVE : A curve of flow vs. Total Head for a specific pump model and impeller  
diameter.  
PERSONAL COMPUTER (PC) : Any general–purpose computer whose size, capabilities, and original  
sales price make it useful for individuals, and which is intended to be operated directly by an end–user  
with no intervening computer operator.  
PERSONAL PROTECTIVE EQUIPMENT (PPE) : Personal protective equipment refers to protective  
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Chapter 25. Glossary  
clothing, helmets, goggles or other garments or equipment designed to protect the wearer’s body from  
injury by blunt impacts, electrical hazards, heat, chemicals and infection, for job–related occupational  
safety and health purposes.  
PH : The intensity of the basic or acid condition of a liquid.  
PH VALUE : A convenient method of expressing small differences in the acidity or alkalinity of solutions.  
Neutrality = pH 7.1; lower values indicate increasing acidity, higher values indicate increasing alkalinity.  
PH VALUE : A convenient method of expressing small differences in the acidity or alkalinity of solutions.  
Neutrality = pH 7.1; lower values indicate increasing acidity, higher values indicate increasing alkalinity.  
PHOTOGRAPHIC INSPECTIONS : A method of obtaining photographs of a pipeline by pulling a  
time–lapse motion picture camera through the line.  
PHOTOSYNTHESIS : A complex process in all green plants that contain chlorophyll. The process uses  
sunlight as energy to convert carbon dioxide into plant growth. As a by–product oxygen is released.  
PIG : Refers to a poly pig which is a bullet–shaped device made of hard rubber or similar material.  
PIPE CAPACITY : In a gravity–flow sewer system, pipe capacity is the total amount in gallons a pipe is  
able to pass in a specific time period.  
PIPE CLEANING : Removing grease, grit, roots and other debris from a pipe run by means of one of the  
hydraulic cleaning methods.  
PIPE DIAMETER : The nominal or commercially designated inside diameter of a pipe, unless otherwise  
stated.  
PIPE DISPLACEMENT : The cubic inches of soil or water displaced by one foot or one section of pipe.  
PIPE FRICTION LOSS : The positive head loss from the friction resistance between the pipe walls and  
the moving liquid.  
PIPE GRADE : The angle of a sewer or a single section of a sewer as installed. Usually expressed in a  
percentage figure to indicate the drop in feet or tenths of a foot per hundred feet. For example, 0.5 percent  
grade means a drop of one–half foot per 100 feet of length.  
PIPE JOINT : A place where two sections of pipe are coupled or joined together.  
PIPE JOINT SEAL : (1) The tightness or lack of leakage at a pipe joint. (2) The method of sealing a pipe  
coupling.  
PIPE LINER : A plastic liner pulled or pushed into a pipe to eliminate excessive infiltration or exfiltration.  
Other solutions to the problem of infiltration/exfiltration are the use of cement grouting or replacement of  
damaged pipe.  
PIPE PLUG : (1) A temporary plug placed in a sewer pipe to stop a flow while repair work is being ac-  
complished or other functions are performed. (2) In construction of a new sewer system, service saddles  
are sometimes installed before a building or a building lateral is in existence. Under such circumstances, a  
plug will be placed in the off–lead of the saddle of a “Y.”  
PIPE RODDING : A method of opening a plugged or blocked pipe by pushing a steel rod or snake, or  
pulling same, through the pipe with a tool attached to the end of the rod or snake. Rotating the rod or  
snake with a tool attached increases effectiveness. 234  
PIPE ROUGHNESS : A measurement of the average height of peaks producing roughness on the internal  
surface of pipes. Roughness is measured in many locations, and is usually defined in micro–inches RMS  
513  
(root mean square).  
PIPE RUN : (1) The length of sewer pipe reaching from one manhole to the next. (2) Any length of pipe,  
generally assumed to be in a straight line.  
PIPE SECTION : A single length of pipe between two joints or couplers.  
PIPING & INSTRUMENTATION DIAGRAM (P&ID) : A diagram in the process industry that shows the  
piping of the process flow together with the installed equipment and instrumentation.  
PISTON EFFECT : The sealing principle involved in utilizing line pressure to effect a seal across the  
floating seats of some valves.  
PISTON PUMP : The piston pump is a positive displacement type of pump. As the piston is pulled back  
it draws in the fluid, and then as it’s pushed forward it pushes the liquid out. A piston pump can have up  
to four pistons depending on the application. They should only be used for clear liquids as any solids  
and/or abrasives in the fluid can damage the pump. Piston pumps are for low flow, high head applications.  
Frequently used for high–accuracy metering applications.  
PLAN : A drawing showing the TOP view of sewers, manholes and streets. Also means approved contract  
drawings, town standards, working drawings, detail sheets or exact reproductions thereof, which show the  
location, character, dimensions and details of the work to be done.  
PLUG : The rotating closure element of a plug valve. Also a threaded fitting used to close off and seal an  
opening into a pressure containing chamber, e.g., pipe plug.  
PLUG VALVE : A quarter turn valve whose closure element is usually a tapered plug having a rectangular  
port.  
PNEUMATIC EJECTOR : A device for raising wastewater, sludge or other liquid by compressed air.  
The liquid is alternately admitted through an inward–swinging check valve into the bottom of an airtight  
pot. When the pot is filled compressed air is applied to the top of the liquid. The compressed air forces  
the inlet valve closed and forces the liquid in the pot through an outward–swinging check valve, thus  
emptying the pot.  
POLISHING POND : A final lagoon cell after other treatment which completes the treatment, or "pol-  
ishes" the effluent.  
POLY PAK STEM SEAL : An O–ring energized lip–seal which replaces O–ring stem seals in certain gate  
valves. Also used for stem seals in some ball valves.  
POLYELECTROLYTES : Synthetic chemicals used as a coagulant aid.Primary Waste Treatment – Me-  
chanical separation of solids, grease, and scum from wastewater. With the aid of flocculating agents,  
primary treatment can eliminate 50 to 65 POLYMER : Polymers are used with other chemical coagulants  
to aid in binding small suspended particles to larger chemical flocs for their removal from water.  
POLYMER : Polymers are used with other chemical coagulants to aid in binding small suspended parti-  
cles to larger chemical flocs for their removal from water.  
POLYPHOSPHATE : A large compound formed of several orthophosphate molecules connected by  
phosphate–storing microorganisms.  
POLYVINYL CHLORIDE : The most common material used for wastewater piping. It is a type of plas-  
tic.  
PONDING : A condition occurring on trickling filters when the hollow spaces (voids) become plugged to  
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Chapter 25. Glossary  
the extent that water passage through the filter is inadequate. Ponding may be the result of excessive slime  
growths, trash, or media breakdown.  
PONDING : A condition occurring on trickling filters when the hollow spaces (voids) become plugged to  
the extent that water passage through the filter is inadequate. Ponding may be the result of excessive slime  
growths, trash, or media breakdown.  
POPULATION EQUIVALENT (HYDRAULIC) : A flow of 100 gallons per day is the hydraulic or flow  
equivalent to the contribution or flow from one person. Population equivalent = 100 GPCD or gallons per  
capita per day.  
POPULATION EQUIVALENT : An average BOD contribution by each person to a domestic sewage. The  
accepted population equivalent is 0.17 pounds of BOD per person per day.  
PORCUPINE : A sewer cleaning tool the same diameter as the pipe being cleaned. The tool is a steel  
cylinder having solid ends with eyes cast in them to which a cable can be attached and pulled by a winch.  
Many short pieces of cable or bristles protrude from the cylinder to form a round brush.  
POTABLE WATER : Water fit for human consumption.  
POTENTIAL ENERGY : A thermodynamic property. The energy associated with the mass and height of  
a body above a reference plane.  
POTENTIAL HYDROGEN : A measurement of how acidic or basic wastewater is on a scale of 0 to 14.  
POUNDS PER SQUARE INCH : A measurement of pressure. It’s often used when discussing physical  
wastewater treatment technologies involving filtration, but is also used with pumps. Filtration system PSI  
can indicate when it’s time to backwash or change a filter.  
POWDER PUMP : Powder pumps are normally of the Air Operated Diaphragm type and really can pump  
just powder and powder like materials such as flour and other fine grained, low bulk, dry–density powders  
in a dust free operation.  
POWER OPERATOR : Powered valve operators are of the following general types.. Electric Motor,  
Pneumatic or Hydraulic Motor, Pheumatic or Hydraulic Cylinder. Operators can either be adapted directly  
to the valve stem or side mounted on existing gear or scotch–yoke operators.  
POWER RODDER : A sewer cleaning machine fitted with auger rods which are inserted in a sewer line to  
dislodge and cut roots and debris.  
PPM : Abbreviation for Parts Per Million. See MILLIGRAMS PER LITER (mg/L)  
PRE: CLEANING – Sewer line cleaning, commonly done by high–velocity cleaners, that is done prior to  
the TV inspection of a pipeline to remove grease, slime, and grit to allow for a clearer and more accurate  
identification of defects and problems. 235  
PRECIPITATE : (1) An insoluble, finely divided substance which is a product of a chemical reaction  
within a liquid. (2) The separation from solution of an insoluble substance.  
PRECIPITATION : (1) The total measurable supply of water received directly from clouds as rain, snow,  
hail, or sleet; usually expressed as depth in a day, month, or year, and designated as daily, monthly, or  
annual precipitation. (2) The process by which atmospheric discharged onto a land or water surfaces. (3)  
The separation (of a substance) out in solid form from a solution, as by the use of a reagent. moisture is  
PRELIMINARY TREATMENT : The removal of rocks, rags, sand, eggshells, and similar materials which  
may hinder the operation of a treatment plant. Preliminary treatment is accomplished by using equipment  
515  
such as bar screens and grit removal systems.  
PRESSURE : The application of external or internal forces to a body producing tension or compression  
within the body. This tension divided by a surface is called pressure.  
PRESSURE DROP : Referring to the loss of pressure between two points in a pipeline system. Generally,  
this occurs because of pipe friction loss or differences in elevation between the two points.  
PREVENTIVE MAINTENANCE : Crews assigned the task of cleaning sewers (for example, balling  
or high–velocity cleaning crews) to prevent stoppages and odor complaints. Preventive maintenance is  
performing the most effective cleaning procedure, in the area where it is most needed, at the proper time  
in order to prevent failures and emergency situations.  
PRIMARY CELL : The first cell in a series, generally receiving raw wastewater.  
PRIMARY CONTAMINANTS : The contaminants identified by the EPA as harmful to human health. In  
order to protect public health, the primary contaminants must not exceed certain specified levels known as  
Maximum Contaminant Levels (MCL).  
PRIMARY TREATMENT (ALSO KNOWN AS SEDIMENTATION) : A wastewater treatment process  
that takes place in a rectangular or circular tank and allows those substances in wastewater that readily  
settle or float o be separated from the water being treated.  
PRIMARY WASTE TREATMENT : Mechanical separation of solids, grease, and scum from wastewater.  
With the aid of flocculating agents, primary treatment can eliminate 50 PROCESS AND INSTRUMEN-  
TATION DIAGRAM : An engineering drawing for a wastewater treatment system. It’s a schematic flow  
diagram that shows the relationship between different instrumentation and equipment.  
PROCESS CONTROL : An engineering discipline that deals with architectures, mechanisms and algo-  
rithms for maintaining the output of a specific process within a desired range.  
PROCESS FLOW DIAGRAM : A diagram commonly used in engineering to indicate the general flow of  
plant processes and equipment. The PFD displays the relationship between major equipment of a plant  
facility and does not show minor details.  
PROCESS IN WHICH CAVITIES OR BUBBLES FORM IN THE FLUID LOW: pressure area and col-  
lapse in a higher pressure area of the pump – causing noise, damage to the pump, and loss of efficiency  
because it distorts the flow pattern. Occurs in centrifugal pumps when NPSHa < NPSH. A properly de-  
signed system and a properly sized pump will prevent cavitation.  
PROFILE : A drawing showing the SIDE view of sewers and manholes.  
PROGRAMMABLE LOGIC CONTROLLER (PLC) : A microprocessor–based system which provides  
plant automation by monitoring sensors and controlling actuators and equipment in real time.  
PROGRESSIVE CAVITY PUMP : A pump that uses a stator and rotor in a screw shape. The rotor turning  
inside the stator causes cavities to move in the direction of flow. Progressive Cavity Pumps cannot run dry  
for any amount of time.  
PROOF PPRESSURE : A hydrostatic test pressure, usually 1 ½ times the rated working pressure, applied  
to an assembled valve to verify the structural integrity of the pressure containing parts. Synonymous with  
hydrostatic shell test. (Table 5.1, API–6D).  
PROPELLER PUMP : Propeller pumps are similar to other centrifugal impeller pumps, but the fluid being  
pumped is not sent in a circular path. Rather, it proceeds more or less in a straight direction up to the  
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Chapter 25. Glossary  
discharge. The motor sits above the discharge shaft. The propeller can be placed below the surface of the  
liquid, where it will always be primed. Propeller pumps are generally low–speed but low heads. They can  
be quite large, measuring over a dozen feet in diameter and moving over 50,000 gallons per minute. Some  
have adjustable–pitch blades.  
PROTECTIVE SLEEVES : A circular "pipe like" sleeve inserted in place of the ball and seats of a top  
entry ball valve. This protective sleeve remains in place inside the valve during valve installation and  
ultimate pigging of a pipeline to clear debris from the line before placing the pipeline into service. Once  
the pipeline has been purged of all debris, the protective sleeve is removed entirely from the ball valve  
cavity and operating trim (i.e. ball and seats) is then installed for normal service conditions.  
PUBLIC SEWER : means a sewer in which all owners of abutting properties have equal rights and is  
controlled by acting as Sewer Commissioners, and maintained by the Public Works Superintendent.  
PUBLICLY OWNED TREATMENT WORK : A term to describe a city or municipal sewage treatment  
facility. Since most industries discharge wastewater to these facilities, they’re typically regulated by these  
POTWs.  
PUMP : A mechanical device for causing flow, for raising or lifting water or other fluid, or for applying  
pressure to fluids.  
PUMP CONTROL VALVE : A ball valve that is not meant for on–off service, but whose specific function  
is to control flow and prevent cavitation in pumps on liquid pipelines.  
PUMP IMPELLER : The moving element in a centrifugal pump that drives the fluid.  
PUMP PIT : A dry well, chamber or room below ground level in which a pump is located.  
PUMP STATION : Installation of pumps to lift wastewater to a higher elevation in places where flat land  
would require excessively deep sewer trenches. Also used to raise wastewater from areas too low to drain  
into available collection lines. These stations may be equipped with  
RACHET DRIVE : A shaft or valve that is operated by means of a ratchet mechanism. The ratchet deliv-  
ers an intermittent stepped rotation through a gear in one direction only.  
RAW WASTEWATER : Wastewater before it receives any treatment.  
REACTOR : A tank where a wastewater stream is mixed with bacterial sludge and biochemical reactions  
occur.  
RECEIVING WATER : A stream, river, lake, ocean or other surface or groundwater into which treated or  
untreated wastewater is discharged.  
RECIRCULATION : The return of part of the effluent from a treatment process to the incoming flow.  
REDUCED PORT : A valve port opening that is smaller than the line size or the valve end connection  
size.  
REGULAR PORT VALVE : A term usually applied to plug valves. The "regular" port of such a valve  
is customarily about 40 REGULATOR : A throttling valve which exercises automatic control over some  
variable (usually pressure). Not an on–off valve.  
RELIEF VALVE : A quick acting, spring loaded valve that opens (relieves) when the pressure exceeds the  
spring setting. Often installed on the body cavity of ball and gate valves to relieve thermal overpressure in  
liquid services.  
REMOTE CONTROL : The operation of a valve or other flow control device from a point at a distance  
517  
from the device being controlled. Can be accomplished by electrical, pneumatic or hydraulic means.  
RESILIENT SEAT : A valve seat containing a soft seal, such as an o–ring, to assure tight shut–off.  
RETENTION : (1) That part of the precipitation falling on a drainage area which does not escape as  
surface stream flow during a given period. It is the difference between total precipitation and total runoff  
during the period, and represents evaporation, transpiration, subsurface leakage, infiltration, and when  
short periods are considered, temporary surface or underground storage on the area. (2) The delay or  
holding of the flow of water and water carried wastes in a pipe system. This can be due to a restriction in  
the pipe, a stoppage or a dip. Also, the time water is held or stored in a basin or wet well.  
RETENTION TIME : The time water, sludge or solids are retained or held in a clarifier or sedimentation  
tank.  
RETURN ACTIVATED SLUDGE : Activated return sludge is normally returned continuously to the  
aeration tank. Recycling of activated sludge back to the aeration tank provides bacteria for incoming  
wastewater. Its should be brown in color with no obnoxious odor and is often also returned in small por-  
tions to the primary settling tanks to aid sedimentation. Settled activated sludge is generally thinner than  
raw sludge. Some activated sludge will be wasted to prevent excessive solids build up.  
RETURN ACTIVATED SLUDGE SOLIDS (RASS) : The concentration of suspended solids in the sludge  
flow being returned from the settling tank to head of the aeration tank.  
RETURN SLUDGE : Settled activated sludge returned to mix with incoming raw or primary settled  
wastewater. When the return sludge rate in the activated sludge process is too low, there will be insuffi-  
cient organisms to meet the waste load entering the aerator.  
REVERSE OSMOSIS : Water purification technology that uses a semipermeable membrane to remove  
dissolved solids, molecules and larger particles from water. Applied pressure is used to overcome osmotic  
pressure and produce high purity water. Reverse osmosis is used to produce ultra pure water for a variety  
of applications.  
RIM PULL : The force required at the edge of the handwheel to generate the required torque at the center  
of the handwheel.  
RIP: RAP – Erosion control by placement of large rocks along an embankment.  
RISING SLUDGE : Rising sludge occurs in the secondary clarifiers of activated sludge plants when the  
sludge settles to the bottom of the clarifier, is compacted, and then starts to rise to the surface, usually as a  
result of denitrification.  
RISING STEM : A valve stem which rises as the valve is opened. A valve stem with threads arranged  
so that as the stem turns, the threads engage a stationary threaded area and lift the stem along with the  
closure element attached to it.  
RISING STEM BALL VALVE : A single seated ball valve that is designed to seal by using the valves  
stem to mechanically wedge the valves ball into a stationary seat effecting a bubble tight seal. The valves  
stem operates througha guide sleeve assembly that guides the stem through a quarter turn of rotation as the  
stem is raised or lowered by a handwheel (or actuator). The mechanical action of the stem moves the ball  
away fromthe seat prior to the 90° rotation of the ball. This design provides lower operating torques and  
longer seat life while assuring bubble tight shut off.  
ROD (SEWER) : A light metal rod, three to five feet long with a coupling at each end. Rods are joined  
and pushed into a sewer to dislodge obstructions.  
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Chapter 25. Glossary  
ROD GUIDE : A bent pipe inserted in a manhole to guide hand and power rods into collection lines so the  
rods can dislodge obstructions.  
RODDING MACHINE : A machine designed to feed a rod into a pipe while rotating the rod. 236  
RODDING TOOLS : Special tools attached to the end of a rod or snake to accomplish various results in  
pipe rodding.  
ROOF LEADER : A downspout or pipe installed to drain a roof gutter to a storm drain or other means of  
disposal.  
ROOT MOP : When roots from plant life enter a sewer system, the roots frequently branch to form a  
growth that resembles a string mop.  
ROOT SEWER : Any part of a root system of a plant or tree that enters a collection system.  
ROTARY GEAR PUMP : A Gear pump uses the meshing of gears to pump fluid by displacement. They  
are one of the most common types of pumps for hydraulic fluid power applications. Gear pumps are also  
widely used in chemical installations to pump fluid with a certain viscosity. There are two main variations;  
external gear pumps which use two external spur gears and internal gear pumps which use an external and  
an internal spur gear. Gear pumps are fixed displacement, meaning they pump a constant amount of fluid  
for each revolution. Some gear pumps are designed to function as either a motor or pump.  
ROTATING BIOLOGICAL CONTACTOR : A biological treatment technology most often used in city  
treatment systems to reduce BOD.  
ROTIFER : A form of microscopic animal that feeds on algae and bacteria. The free swimming protozoa  
are common in lagoons. Rotifers require aerobic conditions.  
SADDLE : A fitting mounted on a pipe for attaching a new connection. This device makes a tight seal  
against the main pipe by use of a clamp, adhesive, or gasket and prevents the service pipe from protruding  
into the main.  
SADDLE CONNECTION : A building service connection made to a sewer main with a device called a  
saddle.  
SAFETY VALVE : A quick opening, pop action valve used for fast relief of excessive pressure.  
SAND TRAP : A device which can be placed in the outlet of a manhole to cause a settling pond to de-  
velop in the manhole invert, thus trapping sand, rocks and similar debris heavier than water. Also may be  
installed in outlets from car wash areas.  
SANITARY COLLECTION SYSTEM : The pipe system for collecting and carrying liquid and liq-  
uid–carried wastes from domestic sources to a wastewater treatment plant.  
SANITARY PUMP : Sanitary pumps describe the materials used for construction of how a pump is built  
and if they meet specific criteria set forth by certifying agencies. Typical describing words are “FDA  
Compliant”, “Food Grade” and “CIP (Clean in Place)” and EHEDG. Sanitary pumps are normally built  
from stainless steel, PTFE, EPDM and other “clean” materials.  
SANITARY SEWER : A pipe or conduit (sewer) intended to carry wastewater or waterborne wastes  
from homes, businesses, and industries to the POTW. Storm water runoff or unpolluted water should be  
collected and transported in a separate system of pipes or conduits (storm sewers) to natural water courses.  
SATURATION : Oxygen saturation is the concentration of free dissolved oxygen in water that is in equi-  
librium with atmospheric oxygen. It is measured in milligrams per liter (mg/l). It varies with both temper-  
519  
ature and atmospheric pressure.  
SCOOTER : A sewer cleaning tool whose cleansing action depends on the development of high water  
velocity around the outside edge of a circular shield. The metal shield is rimmed with a rubber coating  
and is attached to a framework on wheels (like a child’s scooter). The angle of the shield is controlled by a  
chain–spring system which regulates the head of water behind the scooter and thus the cleansing velocity  
of the water flowing around the shield.  
SCOTCH YOKE OPERATOR (USED ON QUARTER TURN VALVES) : A quarter turn operator using  
a scotch yoke mechanism rather than gears. The "Scotch Yoke" has a torque output at the beginning and  
ending of its stroke that is generally twice the magnitude oft he torque output in the center of its stroke.  
SCREEN : A device used to retain or remove suspended or floating objects in wastewater. The screen  
has openings that are generally uniform in size. It retains or removes objects larger than the openings. A  
screen may consist of bars, rods, wires, gratings, wire mesh, or perforated plates.  
SCREEN : A device used to retain or remove suspended or floating objects in wastewater. The screen  
has openings that are generally uniform in size. It retains or removes objects larger than the openings. A  
screen may consist of bars, rods, wires, gratings, wire mesh, or perforated plates.  
SCREW CENTRIFUGAL PUMP : A pump which uses an open channel impeller with a screw shape.  
These pumps are ideal for sludges, large/stringy solids laden fluids, shear sensitive fluids and delicate or  
highly abrasive materials. Offering true non–clog performance and a steep head curve make these pumps  
ideal for wastewater and other sludge applications. They also handle solids more gently than other pumps.  
They are specifically used for transporting live fish without harm and delicate foodstuffs without bruising.  
SCUM : (1) A layer or film of foreign matter (such as grease, oil) that has risen to the surface of water or  
wastewater. (2) A residue deposited on the ledge of a sewer, channel, or wet well at the water surface. (3)  
A mass of solid matter that floats on the surface.  
SEAT : That part of a valve against which the closure element (gate, ball) effects a tight shut–off. In many  
ball valves and gate valves, it is a floating member containing a soft seating element (usually an o–ring).  
SECONDARY : The second in a series of cells.  
SECONDARY CONTAMINANTS : Contaminants in drinking water that are not harmful to human health  
but are unpleasant. Secondary contaminants include substances that cause unpleasant tastes and odors  
or color the water. A Recommended Maximum Level (RCM) has been set for each of the secondary  
contaminants in order to make sure the water is pleasant to drink. 237  
SECONDARY TREATMENT : A wastewater treatment process used to convert dissolved or suspended  
materials into a form more readily separated from wastewater. Usually follows primary sedimentation  
treatment and uses biological processes to convert wastes to solids that settle in secondary clarifiers. Also  
occurs in lagoon systems.  
SECONDARY WASTE TREATMENT : Processing by various types of systems that employ aeration and  
biological oxidation stages to decompose dissolved and colloidal organic contaminants (inorganic plant  
nutrients may also be partially removed).  
SEDIMENT : Solid material settled from suspension in a liquid.  
SEDIMENTATION : The process of settling and depositing of suspended matter carried by wastewater.  
Sedimentation usually occurs by gravity when the velocity of the wastewater is reduced below the point at  
which it can transport the suspended material.  
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Chapter 25. Glossary  
SEDIMENTATION TANKS : Provide a period of quiescence during which suspended waste material  
settles to the bottom of the tank and is scraped into a hopper and pumped out for disposal. During this  
period, floatable solids (fats, oils) rise to the surface of te tank and are skimmed off into scum pipes for  
disposal.  
SELECT BACKFILL : Material used in backfilling of an excavation, selected for desirable compaction or  
other characteristics.  
SELECT BEDDING : Material used to provide a bedding or foundation for pipes or other underground  
structures. This material is of specified quality for desirable bedding or other characteristics and is often  
imported from a different location.  
SELF RELIEVING : The process whereby excessive internal body pressure, in some valves, is automat-  
ically relieved either into the upstream or downstream line by forcing the seats away from the closure  
element.  
SELF: Priming Pump – Self–priming pumps are centrifugal pumps with an abnormally large and specially  
shaped volute. The purpose of the large volute is to allow the pump to pull or “lift” liquid up to the im-  
peller. Initially the pump volute (casing), must be filled with liquid manually to “pre–prime” the pump. As  
the pump starts it pumps out the liquid that was manually put into it while also drawing up the air in the  
suction pipe along with pulling up the liquid to be pumped. As the lifted liquid enters the volute the final  
volume of air is pumped out of the discharge and through an air release valve. Once the liquid hits the air  
valve, it closes and the pump now operates as a standard centrifugal pump.  
SEPTIC : A condition that exists when there is no dissolved oxygen (see anaerobic). Anaerobic bacteria  
and other microorganisms continue to use parts of the waste for food, but produce foul odors and black  
colored water. The waste in the common septic tank is typical of this condition.  
SEQUENCING BATCH REACTORS : A biological treatment technology based on the activated sludge  
process. It is sometimes used in small municipalities and at food processing facilities who discharge  
directly to streams or rivers.  
SERVICE ROOT : A root entering the sewer system in a service line and growing down the pipe and into  
the sewer main.  
SETTLEABILITY : A process–control test used to evaluate the settling characteristics of activated sludge.  
Reading taken at 30 to 60 minutes are used to calculate the settled sludge volume (SSV) and the sludge  
volume index (SVI).  
SETTLED SLUDGE VOLUME (SSV) : The volume in percent occupied by an activated sludge sample  
after 30 to 60 minutes of settling. Normally written as SSV with a subscript to indicate the time of the  
reading used for calculation (SSV60) or (SSV30).  
SEWAGE : Largely the water supply from a community after it has been fouled by various uses. From  
the standpoint of course, it may be a combination of the liquid or water–carried wastes from residences,  
business buildings, and institutions, together with those from industrial establishments, and with such  
ground water, surface water, and storm water as may be present.  
SEWER : A pipe or conduit that carries wastewater or drainage water.  
SEWER BALL : A spirally grooved, inflatable, semi–hard rubber ball designed for hydraulic cleaning of  
sewer pipes.  
SEWER CLEANOUT : A capped opening in a sewer main that allows access to the pipes for rodding and  
521  
cleaning. Usually such cleanouts are located at terminal pipe ends or beyond terminal manholes.  
SEWER GAS : (1) Gas in collection lines (sewers) that results from the decomposition of organic matter  
in the wastewater. When testing for gases found in sewers, test for lack of oxygen and also for explosive  
and toxic gases. (2) Any gas present in the wastewater collection system, even though it is from such  
sources as gas mains, gasoline, and cleaning fluid.  
SEWER JACK : A device placed in manholes which supports a yoke or pulley that keeps wires or cables  
from rubbing against the inlet or outlet of a sewer.  
SEWER MAIN : A sewer pipe to which building laterals are connected.  
SEWER USE DISCHARGE PERMIT : Permit required or issued jointly by the Authority and a Munici-  
pality for the discharge of industrial waste. 238  
SEWERAGE : System of piping with appurtenances for collecting, moving and treating wastewater from  
source to discharge.  
SEWERAGE SYSTEM : Any device, equipment or works used in the transportation, pumping, storage,  
treatment, recycling, and reclamation of Wastewater and Industrial Wastes.  
SEWERS : A system of pipes used for collecting domestic and industrial waste, as well as storm water  
run–off. Lateral sewers connect homes and industries to trunk sewers, which channel waste into intercep-  
tor sewers carry only domestic and industrial wastewater. Storm sewers carry only storm water run–off.  
Combined sewers carry both.  
SHORING : Material such as boards, planks or plates, and jacks used to hold back soil around trenches  
and to protect workers in a trench from cave–ins.  
SHORT CIRCUITING : A condition that occurs in tanks or basins when some of the water travels faster  
than the rest of the flowing water. This is usually undesirable since it may result in shorter contact, reac-  
tion, or settling times in comparison with the theoretical (calculated) or presumed detention times.  
SHORT GATE : A gate valve whose seat rings contact the gate only in the closed position. Such valves  
are not through conduit, as the gate is completely withdrawn from the flow area in the open position.  
SHORT PATTERN VALVE : A valve whose face–to–face dimension is less than the API–6D standard.  
SHUT: off Head – The Total Head corresponding to zero flow on the pump performance curve.  
SHUT: OFF VALVE – A valve designed only for on/off service. Not a throttling valve. Sometimes re-  
ferred to as a "block valve."  
SILT DENSITY INDEX : Measurement of silt, colloids, bacteria and other foulants of Reverse Osmo-  
sis (RO) membranes. SDI is used to help determine the suitability of water or other liquids for the RO  
process.  
SILTING : Silting takes place when the pressure of infiltrating waters is great enough to carry silt, sand  
and other small particles from the soil into the sewer system. Where lower velocities are present in the  
sewer pipes, settling of these materials results in silting of the sewer system.  
SILVER : A metal element regulated by wastewater discharge permits and common in metal finishing  
wastewater.  
SIPHON : Is a system of piping or tubing where the exit point is lower than the entry point.  
SLAB GATE : A gate having flat, finely finished, parallel faces – as opposed to a wedge gate. Such a  
closure element slides across the seats and does not depend on stem force to achieve tight shut off.  
522  
Chapter 25. Glossary  
SLAM RETARDER : A device designed to prevent the clapper of a check valve from slamming as it  
closes upon flow reversal. Hydraulic damping cylinders, rotary vanes, and torsional springs are all used  
for this purpose.  
SLEEVE : A pipe fitting for joining two pipes of the same nominal diameter in a straight line.  
SLIPLINING : A sewer rehabilitation technique accomplished by inserting flexible polyethylene pipe into  
an existing deteriorated sewer.  
SLOPE : The slope or inclination of a sewer trench excavation is the ratio of the vertical distance to the  
horizontal distance or “rise over run.” The inclination of a trench bottom or a trench sidewall, expressed as  
a ratio of vertical distance to the horizontal distance. For example, a 3:1 slope shall rise or fall 3’ vertical  
feet in a distance of 1’ horizontal foot.  
SLUDGE : The accumulated suspended solids of sewage deposited in tanks or basins.  
SLUDGE : The accumulated suspended solids of sewage deposited in tanks or basins.  
SLUDGE : (1) The settleable solids separated from liquids during processing. (2) The deposits of foreign  
material on the bottoms of streams or other bodies of water.  
SLUDGE AGE : In the activated sludge process, a measure of the length of time a particle of suspended  
solids has been undergoing aeration, expressed in day. It is usually computed by dividing the weight of the  
suspended solids in the aeration tank by the weight of excess activated sludge discharged from the system  
per day.  
SLUDGE DIGESTION : The process of changing organic matter in sludge into a gas or liquid or a more  
stable solid form. These changes take place as microorganisms feed on sludge in anaerobic (more com-  
mon) or aerobic digesters.  
SLUDGE DIGESTION : The process of changing organic matter in sludge into a gas or liquid or a more  
stable solid form. These changes take place as microorganisms feed on sludge in anaerobic (more com-  
mon) or aerobic digesters.  
SLUDGE DIGESTION : The purpose of sludge digestion is to separate the liquid from the solids to  
facilitate drying. The proper pH range for digested sludge is 6.8 – 7.2.  
SLUDGE INDEX : Properly called sludge volume index (SVI). It is the volume in millimeters occupied  
by 1 g of activated sludge after settling of the aerated liquid for 30 minutes.  
SLUDGE JUDGE : A clear tubular device used to measure sludge depth in clarifiers or other tanks.  
SLUDGE LOADING RATE : The weight of wet bio–solids fed to the reactor per square foot of reactor  
bed area per hour (lb./ft2/H).  
SLUDGE REAERATION : The continuous aeration of sludge after initial aeration for the purpose of  
improving or maintaining its condition.  
SLUDGE VOLUME INDEX (SVI) : A process–control calculation used to evaluate the settling quality of  
activated sludge. Requires SSV30 and mixed liquor suspended solids test results to calculate.  
SLURRY : Slurry is defined as a suspension of solids in a liquid. Typically, the liquid is water and the  
solids can be anything from soft materials such as sewage and food processing waste (potato skins, fish  
parts) to abrasive solids like sand, fly ash and coal. Keep in mind a typical centrifugal pump really can’t  
handle more than 3 SMOKE TEST : A method of blowing smoke into a closed–off section of a sewer  
system to locate sources of surface inflow.  
523  
SNAKE : A stiff but flexible cable that is inserted into sewers to clear stoppages.  
SOAP CAKE OR SOAP BUILDUP : A combination of detergents and greases that accumulate in sewer  
systems, build up over a period of time, and may cause severe flow restrictions.  
SOIL POLLUTION : The leakage (exfiltration) of raw wastewater into the soil or ground area around a  
sewer pipe.  
SOLENOID VALVE : A small electrically operated valve used in the control piping of powered by hy-  
draulic or pneumatic cylinder operators.  
SOLIDS FEED RATE : The dry solids fed to a centrifuge.  
SOLIDS HANDLING PUMP : Many types of pumps can be used for solids handling. The size and con-  
centration of solids in the fluid will determine the best type of pump for the application. For sewage  
applications such as lift stations where the solids concentration does not exceed 3 SOLIDS LOADING  
(BELT FILTER PRESS) : The feed solids to the belt filter on a dry weight basis including chemicals per  
unit time.  
SOLIDS LOADING RATE (DRYING BEDS) : The weight of solids on a dry weight basis applied annu-  
ally per square foot of drying bed area.  
SOLIDS RECOVERY (CENTRIFUGE) : The ratio of cake solids to feed solids for equal sampling times.  
It can be calculated with suspended solids and flow data or with only suspended solids data. The cenrate  
solids must be corrected if chemicals are fed to the centrifuge.  
SOLUBLE BOD : Soluble BOD is the BOD of water that has been filtered in the standard suspended  
solids test.  
SOLUTION : A liquid mixture of dissolved substances. In a solution it is impossible to see all the sepa-  
rated parts.  
SOUNDING ROD : A T–shaped tool or shaft that is pushed or driven down through the soil to locate  
underground pipes and utility conduits.  
SPECIFIC GRAVITY : The ratio of the density of a fluid to that of water at standard conditions  
SPECIFIC SPEED : A formula that describes the shape of a pump impeller. The higher the specific speed  
the less N.P.S.H. required.  
SPLIT CASE PUMP : The split case is almost synonymous with multi–stage and can be either horizontal  
or vertical. Used where high pressures are needed such as boiler feed.  
SPLITTER BOX : A division box that splits the incoming flow into two or more streams. A device for  
splitting and directing discharge from the head box to two separate points of application.  
SPOIL : Excavated material such as soil from the trench of a sewer.  
SPST : SINGLE–POLE, SINGLE THROW – Refers to the function of an electrical switch often used in  
the control system of electric valve operators.  
SPUR GEAR : The simplest of gears. In a gear set, the input spur gear and output spur gear are aligned  
on parallel shafts. An idler gear may be used to the direction of rotation on the two shafts is in the same  
direction.  
SQUARE OPERATING NUT : A nut, usually 2" x 2", which is attached to a valve stem or the pinion  
shaft of a gear operator allowing use of wrenches to quickly operate the valve.  
524  
Chapter 25. Glossary  
SSIV (SUB SEA ISOLATION VALVE) : A valve used underwater, generally in a manifold that will close  
and isolate a particular pipeline or process in an emergency.  
STABILIZATION : The conversion of biodegradable materials into more stable solids. Stabilization is  
the primary function of wastewater lagoons and treatment plants. Lagoons are often called stabilization  
ponds.  
STATION : A point of reference or location in a pipeline is sometimes called a “station.” As an example, a  
building service is located 51 feet downstream from a manhole could be reported to be at “station 51.”  
STEM : A rod or shaft used to transmit motion from an operator to the closure element of a valve.  
STEM INDICATOR (VPI : VISIBLE POSITION INDICATOR) – A position indicating rod supplied with  
gate valves. It extends from the top of the valve stem and serves to indicate the relative position of the  
gate.  
STEM NUT : A one or two–piece nut which engages the stem threads of a valve and transmits torque  
from an operator to the valve stem.  
STILLING WELL : A well or chamber which is connected to the main flow channel by a small inlet.  
Waves and surges in the main flow stream will not appear in the well due to the small diameter inlet. The  
liquid surface in the well will be quiet, but will follow all of the steady fluctuations of the open channel.  
The liquid level in the well is measured to determine the flow in the main channel.  
STOP COLLAR : The collar on a ball valve which restricts the ball to 90° of rotation from the fully open  
to the fully closed position.  
STOPPAGE : Partial or complete interruption of flow as a result of some obstruction in a sewer. (2) When  
a sewer system becomes plugged and the flow backs up, it is said to have a “stoppage.”  
STORM COLLECTION SYSTEM : A system of gutters, catch basins, yard drains, culverts and pipes  
for the purpose of conducting storm waters from an area, but intended to exclude domestic and industrial  
wastes.  
STORM SEWER : A separate pipe, conduit or open channel (sewer) that carries runoff from storms,  
surface drainage, and street wash, but does not include domestic and industrial wastes.  
STRAIN : The ratio between the absolute displacements of a reference point within a body to a character-  
istic length of the body.  
STRATIFICATION : The formation of indistinct layers of slightly variable density of waters. Often  
caused by warming of the surface with an absence of mixing.  
STRESS : In this case refers to tangential stress or the force between the layers of fluid divided by the  
surface area between the layers.  
STRETCH : Length of sewer from manhole to manhole. 240  
STUFFING BOX : The annular chamber provided around a valve stem in a sealing system into which  
deformable packing is introduced.  
SUBMERSIBLE PUMP : Just like it sounds, these guys operate within the fluid they are pumping. Sub-  
mersible pumps can be either centrifugal or AOD type pumps. The centrifugal versions are common used  
in sewage lift stations, while the AODs are used in chemical transfers.  
SUCKER RODS : Rigid, coupled sewer rods of metal or wood used for clearing stoppages. Usually  
available in 3–ft, 39–in, 4–ft, 5–ft and 6–ft lengths.  
525  
SUCTION HEAD : Condition that occurs when the liquid source is above the centerline of the pump.  
SUCTION LIFT : Condition that occurs when the liquid source is below the centerline of the pump.  
SUCTION STATIC HEAD : The difference in elevation between the liquid level of the source of supply  
and the centerline of the pump. This head also includes any additional head that may be present at the  
suction tank fluid surface.  
SUCTION STATIC LIFT : The same definition as the Suction Static head. This term is only used when  
the pump centerline is above the suction tank fluid surface.  
SUPERNATANT : Liquid removed from settling sludge. Supernatant commonly refers to the liquid  
between the sludge on the bottom and the scum on the surface of an anaerobic digester. The liquid is  
usually returned to the influent wet well or to the primary clarifier.  
SUPERVISORY CONTROL AND DATA ACQUISITION (SCADA) : A computer system for gathering  
and analyzing real time data. SCADA systems are used to monitor and control a plant or equipment in  
industry.  
SURCHARGE : Sewers are surcharged when the supply of water to be carried is greater than the capacity  
of the pipes to carry the flow. The surface of the wastewater in manholes rises above the top of the sewer  
pipe, and the sewer is under pressure or a head, rather than at atmospheric pressure.  
SURCHARGED MANHOLE : A manhole in which the rate of the water entering is greater than the  
capacity of the outlet under gravity flow conditions. When the water in the manhole rises above the top of  
the outlet pipe, the manhole is said to be “surcharged.”  
SURFACE LOADING : Lagoon loading is rated organically in pounds of BOD per acre of surface area  
per day. Northern climates require lower loading rates than warmer areas, because cold weather slows  
down the stabilization processes of microorganisms. Treatment plants clarifiers are rated hydraulically in  
flow (gpd) per surface area (sq ft).  
SURFACTANT : Compounds that lower the surface tension (or interfacial tension) between two liquids  
or between a liquid and a solid. Surfactants may act as detergents, wetting agents, emulsifiers, foaming  
agents, and dispersants.  
SUSPENDED SOLID : Solids that either float on the surface or are suspended in water, wastewater, or  
other liquids, and which are largely removable by laboratory filtering.  
SWAB : A circular sewer cleaning tool almost the same diameter as the pipe being cleaned. As a final  
cleaning procedure after a sewer line has been cleaned with a porcupine, a swab is pulled through the  
sewer and the flushing action of water flowing around the tool cleans the line.  
SWING CHECK VALVE : A check valve in which the closure element is a hinged clapper which swings  
or rotates about a supporting shaft.  
SYSTEM : Systems, as far as pumps are concerned, include all the piping with or without a pump, start-  
ing at the inlet point (often the fluid surface of the suction tank) and ending at the outlet point (often the  
fluid surface of the discharge tank).  
SYSTEM CURVE : Is a plot of flow vs. Total Head that satisfies the system requirements.  
SYSTEM EQUATION : The equation for Total Head vs. flow for a specific system.  
SYSTEM REQUIREMENTS : Friction and system inlet and outlet conditions (i.e. velocity, elevation and  
pressure).  
526  
Chapter 25. Glossary  
TAG LINE : A line, rope or cable that follows equipment through a sewer so that equipment can be pulled  
back out if it encounters an obstruction or becomes stuck. Equipment is pulled forward with a pull line.  
TAP : A small hole in a sewer where a wastewater service line from a building is connected (tapped) into a  
lateral or branch sewer.  
TELEVISION INSPECTION : An inspection of the inside of a sewer pipe made by pulling a closed–circuit  
television camera through the pipe.  
TERMINAL “LAMPHOLES” CLEANOUT : When a manhole is not provided at the upstream end of a  
sewer main, a cleanout is usually provided. This is called a “terminal cleanout.”  
TERTIARY WASTE TREATMENT : Following secondary treatment, the clarified effluent may require  
additional aeration and/or other chemical treatment to destroy bacteria remaining from the secondary  
treating stage, and to increase the content of dissolved oxygen needed for oxidation of the residual BOD.  
Tertiary treatment can also be used to remove nitrogen and phosphorous.  
THROTTLING : The intentional restriction of flow by partially closing or opening a valve. A wide range  
of throttling is accomplished automatically in regulators and control valves.  
THRU: CONDUIT – An expression characterizing valves when in the open position, wherein the bore  
presents a smooth uninterrupted interior surface across seat rings and thru the valve port, thus affording  
minimum pressuredrop. There are no cavities or large gaps in the bore between seat rings and body clo-  
sures or between seat rings and ball/gate. Consequently, there are no areas that can accumulate debris to  
impede pipeline cleaning equipment or restrict the valve’s motion.  
THRUST : The net force applied to a part in a particular direction – e.g., on the end of a valve stem.  
TOP ENTRY : The design of a particular valve or regulator where the unit can be serviced or repaired by  
leaving its body in the line, and its internals can be accessed by removing a top portion of the unit.  
TORQUE : The turning effort required to operate a valve. Usually expressed in "pound–feet" and referred  
to the stem nut, handwheel or operator pinion shaft.  
TORQUE SWITCH : An electrical device on a motor operator which cuts off power to the operator when  
allowable torque is exceeded, thus preventing damage to the valve and/or the operator.  
TORSIONAL SPRING : A coiled spring which exerts a force by twisting about its axis rather than by  
compression or elongation. The spring in a check valve slam retarder which is restrained at one end and  
fastened tot he clappershaft on the other end. As the clapper opens, the spring resists the motion creating a  
closing force. During a rapid decrease in flow rate, the clapper is urged toward the closed position and is  
virtually closed just prior to the instant of actual flow reversal – thus slamming is avoided.  
TOTAL DISSOLVED SOLIDS : Total dissolved solids are inorganic molecules of metals, minerals or  
salts present in water at such a small size that you can’t see them. Because of their very small size, they  
can be difficult to remove with any technology other than fine membrane filtration technologies such as  
Reverse Osmosis (RO).  
TOTAL HEAD / TOTAL DYNAMIC HEAD : The amount of head produced by the pump. Calculated by  
summing the static head, friction head, pressure head, and velocity head.  
TOTAL KJELDAHL NITROGEN (TKN): Is a standard laboratory analysis which measures organically  
bound nitrogen plus ammonia nitrogen.  
TOTAL ORGANIC CARBON : A direct measurement of how much organic matter is in wastewater.  
527  
TOTAL SOLIDS : The total amount of solids in solution and suspension.  
TOTAL STATIC HEAD : The difference between the discharge and suction static head including the  
difference between the surface pressure of the discharge and suction tanks.  
TOTAL SUSPENDED SOLIDS : Visible solids present in wastewater that can be filtered out through  
traditional physical treatment technologies. In the metal finishing industry, for example, FOG (fats, oils  
and grease) and dirt particles might make up part of the total suspended solids.  
TOTAL TOXIC ORGANICS : A wastewater parameter that refers to the entire amount of toxic organic  
compounds present. EPA has developed a specific list of chemicals that are defined as toxic organic  
compounds.  
TOXIC : A substance which is poisonous to a living organism.  
TOXIC : A substance which is poisonous to a living organism.  
TOXIC ORGANIC MANAGEMENT PLAN : A spill plan federally required for specific industries,  
including metal finishing. It outlines what specific toxic organic compounds are used and how they are  
disposed in a manner that prevents discharge to the sewer system.  
TOXICITY : The relative degree of being poisonous or toxic. A condition which may exist in wastes and  
will inhibit or destroy the growth or function of certain organisms.  
TRANSPIRATION : See Evapotranspiration.  
TRICKLING FILTER : An aerobic biological process used as secondary treatment of sewage. Effluent  
from the primary clarifier is distributed over a bed of rocks. As the liquid trickles over the rocks, a bio-  
logical growth on the rocks breaks down the organic matter in the sewage. The effluent is then taken to a  
clarifier to remove biological matter coming from the filter.  
TRIM : Commonly refers to the valve’s working parts and to their materials. Usually includes seat ring  
sealing surfaces, closure element sealing surfaces, stems, and back seats. Trim numbers which specifythe  
materials are defined in API 600 and API 602.  
TRIPLE ECCENTRIC (BUTTERFLY VALVES) : A particular design of a butterfly valve where the stem  
is located behind the disc, below the centerline of the disc, and its cone axis is offset from the centerline of  
the disc. This particular designis capable of a very tight shutoff at temperatures well above 100°F.  
TRUNNION : That part of a ball valve which holds the ball on a fixed vertical axis and about which the  
ball turns. The torque requirement of a trunnion mounted ball valve is significantly less than that for a  
floatingball design.  
TSS : Abbreviation for TOTAL SUSPENDED SOLIDS, a test measuring the amount of filterable solids in  
wastewater.  
TURBID : Having a cloudy or muddy appearance.  
TURBIDITY : The cloudy appearance of water caused by the presence of suspended and colloidal matter.  
A turbidity measurement is used to indicate the clarity of water.  
TURNS TO OPERATE : The number of complete revolutions of a handwheel or the pinion shaft of a gear  
operator required to stroke a valve from fully open to fully closed or vice versa.  
TWO: WAY SPHERE–LOK – A Sphere–Lok with two ports.  
U: CUP (RING–PACKING) – A "U" cross–section ring located on the tail end of certain ball valve seats  
to retain the grease in an emergency seat seal system.  
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Chapter 25. Glossary  
ULTRA FILTRATION : Type of membrane filtration that separates suspended solids and solutes of high  
molecular from a liquid and low molecular weight solutes. The ultra filtration separation process is used  
in industry processes, water treatment and research for purifying and concentrating macromolecular  
solutions.  
ULTRAVIOLET : In some industries, ultraviolet light is used to sterilize water treated wastewater prior to  
reuse or recycling. UV light keeps algae and other bacteria from growing in the recycled wastewater.  
UNION BONNET : A type of valve construction in which the bonnet is held on by a union nut with  
threads on the body.  
UV : Ultraviolet light. UV is useful as a method of disinfection. It leaves no residual and is often used  
where no chlorine residual (or a very low residual) is allowed to be discharged.  
UV DISINFECTION (UVGI) : Ultraviolet germicidal irradiation (UVGI) is a disinfection method that  
uses ultraviolet (UV) light at sufficiently short wavelength to kill microorganisms. It is used in a variety  
of applications, such as food, air and water purification. UVGI utilizes short–wavelength ultraviolet  
radiation (UV–C) that is harmful to microorganisms. It is effective in destroying the nucleic acids in  
these organisms so that their DNA is disrupted by the UV radiation, leaving them unable to perform vital  
cellular functions.  
VALVE : A device used to control the flow of fluid contained in a pipe line.  
VAPOR PRESSURE : The pressure at which a liquid boils at a specified temperature.  
VARIABLE FREQUENCY DRIVE (VFD) : A system for controlling the rotational speed of an alter-  
nating current (AC) electric motor by controlling the frequency of the electrical power supplied to the  
motor.  
VARIABLE FREQUENCY DRIVE : A variable–frequency drive (VFD) is a system for controlling the  
rotational speed of an alternating current (AC) electric motor by controlling the frequency of the electrical  
power supplied to the motor. A variable frequency drive is a specific type of adjustable–speed drive.  
Variable–frequency drives are also known as adjustable–frequency drives (AFD), variable–speed drives  
(VSD), AC drives, micro drives or inverter drives. Since the voltage is varied along with frequency, these  
are sometimes also called VVVF (variable voltage variable frequency) drives. Variable–frequency drives  
are widely used. For example, in water booster stations, pump speed is controlled by the VFD based on  
system demand.  
VARIABLE ORIFICE : A small variable profile valve put in a flow line and used with a pilot to restrict  
the flow into the pilot and make the pilot more or less sensitive to changing conditions.  
VDS : VALVE DATA SHEET – A data sheet defining the minimum level of a valve design, including the  
materials, testing, inspection, and certification requirements.  
VELOCITY HEAD DIFFERENCE : The difference in velocity head between the outlet and inlet of the  
system.  
VENT PLUG : (VENT PLUG ASSEMBLY) – (SAFETY VENT PLUG) – A special pipe plug having a  
small allen–wrench operated vent valve. These special plugs are located at the bottom of most ball valves.  
With the line valve closed (and under pressure) the body cavity pressurecan be vented thru this small valve  
to check tightness of seat seals or to make minor repairs. Having vented the body pressure, the vent plug  
may be removed to blow out debris and foreign material or to flush the body cavity. On some gate valves,  
the vent plug is installed on the bonnet for the sole purpose of venting the body. Such valves have separate  
529  
drain valves.  
VENTURI VALVE : A reduced bore valve. A valve having a bore smaller in diameter than the inlet or  
outlet. For example, an 8"x 6" x 8" ball valve has 8" inlet and outlet connections while the ball and seats  
are 6". The flow through a venture valve will be reduced because of the smaller port. Venturi valves can  
often be economically substituted for plug valves.  
VERTICAL TURBINE PUMP : A vertical turbine pump is a centrifugal type pump, often with multiple  
stages, where the motor is set at ground level and connect via shaft to the pump below. Used as well  
pumps for irrigation, they can also pump from rivers, lakes and other bodies of water.  
VISCOSITY : A property, which measures a fluid’s resistance to movement. The resistance is caused  
by friction between the fluid and the boundary wall and internally by the fluid layers moving at different  
velocities.  
VOLATILE ORGANIC COMPOUNDS : In wastewater, VOCs typically show up as cleaning solvents.  
These chemicals can kill the microbes in POTWs (publicly owned treatment work) if they are discharged  
in large quantities, so they are carefully regulated. They’re challenging to treat because they dissolve in  
water.  
VOLATILE SOLIDS : Those solids in water, wastewater, or other liquids that are lost on ignition of the  
dry solids at 550oC.  
VOLATILE SOLIDS : Those solids in water, wastewater, or other liquids that are lost on ignition of the  
dry solids at 550oC.  
W.O. : WRENCH OPERATED – The operation of a valve by means of a handle or lever. Used on smaller  
size and lower pressure class valves.  
WALL THICKNES : The thickness of the wall of the pressure vessel or valve. For steel valves, minimum  
thickness requirements are defined in ASME B16.34, API 600, and API 602.  
WASTE ACTIVATED SLUDGE : That portion of sludge from the secondary clarifier in the activated  
sludge process that is wasted to avoid a buildup of solids in the system.  
WASTE TREATMENT SLUDGE : A series of tanks, screens, filters, and other processes by which most  
pollutants are removed from water.  
WASTEWATER : The used water and solids from a community that flow to a treatment plant. Storm  
water, surface water, and groundwater infiltration also may be included in the wastewater that enters a  
wastewater treatment plant. The term “sewage” usually refers to household wastes, but this word is being  
replaced by the term “wastewater.  
WASTEWATER TREATMENT PLANT : A city, municipal or industrial facility that’s treating wastewa-  
ter.  
WATER HAMMER : The physical effect, often accompanied by loud banging, produced by pressure  
waves generated within the piping by rapid change of velocity in a liquid system.  
WATER POLLUTION : A general term signifying the introduction into water of micro–organisms, chemi-  
cals, wastes, or sewage which renders the water unfit for it’s intended use.  
WATER QUALITY ACT : Montana’s primary water pollution control legislation that parallels the federal  
CLEAN WATER ACT. It establishes the public policy for Montana to – 1) conserve water resources by  
protecting, maintaining and improving water quality for all its beneficial uses, and 2) provides a compre-  
530  
Chapter 25. Glossary  
hensive program for the prevention, abatement and control of water pollution.  
WATER TREATMENT FACILITY : A city or municipal water treatment facility that’s treating water you  
drink or use in an industrial process.  
WEDGE GATE : A gate whose seating surfaces are inclined to the direction of closing thrust so that  
mechanical force on the stem produces tight contact with the inclined seat rings.  
WEIR : (1) A wall or plate placed in an open channel and used to measure the flow of water. The depth  
of the flow over the weir can be used to calculate the flow rate, or a chart or conversion table may be used.  
(2) A wall or obstruction used to control flow (from settling tanks and clarifiers) to assure a uniform flow  
rate and avoid short–circuiting.  
WELL PUMP : Well pumps are a centrifugal, submersible type of pump used for bringing underground  
water up to the surface for domestic use. They can consist of one or several “stages” depending on the  
well depth and desired discharge pressure. Electrically powered the motor is typically on the bottom of  
the pump with the suction in the middle and the water is pumped upwards through the impeller(s) and  
upwards toward the surface.  
WET OXIDATION : A method of treating or conditioning sludge before the water is removed. Com-  
pressed air is blown into the sludge; the air and sludge mixture is fed into a pressure vessel where the  
organic material is stabilized.  
WET WELL : A compartment or tank in which wastewater is collected. The suction pipe of a pump may  
be connected to the wet well or a submersible pump may be located in the wet well.  
WETTED PARTS : A term used for any part that comes into contact with the process fluid. These parts  
must be checked for chemical compatibility with the process fluid.  
WORK : The energy required to drive the fluid through the system. A measure of a liquid’s resistance to  
flow. Essentially it’s a how thick the liquid is. The viscosity determines the type of pump used, the speed  
it can run at, and with gear pumps, the internal clearances required.  
YOKE : That part of a gate valve which serves as a spacer between the bonnet and the operator or actua-  
tor.  
ZINC : A heavy metal commonly regulated by wastewater permits. It is widely present in all industries  
and can be difficult to treat to low levels through typical physical/chemical treatment technologies. It  
is very important to determine the sources of zinc in process wastewater in order to adequately control  
discharge levels. Other heavy metals include – Arsenic (As), Cadmium (Cd), Chromium(Cr), Copper  
(Cu), Lead (Pb), and Nickel (Ni).  
ZOOGLEAL MASS : Jelly like masses of bacteria found in both the trickling filter and activated sludge  
processes. See also Biomass.